- We did Section 4.3: Linear independent sets; bases today. Suggested problems for Section 4.3: 14, 15, 21, 22, 23, 25, 26, 31, 32, 33, 34
- Suggested problems for Section 4.4: 8, 9, 12, 13, 14, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32
- Suggested problems for Section 4.7: 4-10, 13, 14
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We talked about the following concepts this week:
- The definition of an abstract vector space on page 192 and posted on March 8.
- The definition of a subspace on page 195
- The definition of the span of a set of vectors in the first paragraph of the subsection "A Subspace Spanned by a Set" starting on page 196. This definition in the set-builder notation reads: Let $m$ be a positive integer and let $\mathbf{v}_1,\ldots,\mathbf{v}_m$ be vectors in a vector space $\mathcal{V}.$ The span of vectors $\mathbf{v}_1,\ldots,\mathbf{v}_m$ is defined as \[ \operatorname{Span}\{\mathbf{v}_1,\ldots,\mathbf{v}_m\} = \bigl\{ \alpha_1 \mathbf{v}_1 + \cdots + \alpha_m \mathbf{v}_m : \alpha_1,\ldots,\alpha_m \in \mathbb{R} \bigr\}. \] This is a very important theorem: Theorem 1. $\operatorname{Span}\{\mathbf{v}_1,\ldots,\mathbf{v}_m\}$ is a subspace of $\mathcal{V}.$
- The definition of a linearly independent set on page 210. Next, I will restate this definition as an implication: An indexed set of vectors $\{\mathbf{v}_1,\ldots,\mathbf{v}_m\}$ in a vector space $\mathcal{V}$ is said to be linearly independent if the following implication holds \[ \alpha_1 \mathbf{v}_1 + \cdots + \alpha_m \mathbf{v}_m = \mathbf{0} \quad \text{implies} \quad \alpha_k = 0 \quad \text{for all} \quad k \in \{1,\ldots,m\}. \] There are many other equivalent ways of stating this definition. However, the above statement is the only formal definition which is easiest to use when we need to prove that certain vectors are linearly independent.
- The definition of a basis on page 211.
- Important examples of finite dimensional vector spaces are spaces of polynomials. For $n\in\mathbb{N}$ by $\mathbb{P}_n$ we denote the vector space of all polynomials of degree less or equal to $n.$ The most important step in understanding the vector space $\mathbb{P}_n$ is establishing that the monomials form a basis of this space. I wrote this webpage with a proof which uses only linear algebra. The proof which I give below for $\mathbb{P}_2$ uses calculus. The proof which is given in our textbook uses the Fundamental Theorem of Algebra (which is more difficult to prove.)
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The friendliest among spaces $\mathbb{P}_n$ is the spaces $\mathbb{P}_2$. Below I explore some of the properties of $\mathbb{P}_2$.
- $\mathbb{P}_2$ denotes the vector space of all polynomials of degree less or equal $2.$ That is, in set-builder notation, \[ \mathbb{P}_2 = \bigl\{a_0 + a_1 x +a_2 x^2 \, : \, a_0, a_1, a_2 \in \mathbb{R} \bigr\}. \] Recall that the constant polynomial $1$ is a polynomial in $\mathbb{P}_2$. To get this polynomial in the above set-builder notation we take $a_0 = 1,$ $a_1 = 0,$ and $a_2 = 0.$ To get the polynomial $x$ in the above set-builder notation we take $a_0 = 0,$ $a_1 = 1,$ and $a_2 = 0.$ Similarly, to get the square polynomial $x^2$ in the above set-builder notation we take $a_0 = 0,$ $a_1 = 0,$ and $a_2 = 1.$ Using the concept of the span the above expression for $\mathbb{P}_2$ in set-builder notation can be written using the concept of the span as \[ \mathbb{P}_2 = \operatorname{Span}\bigl\{ 1, x, x^2 \bigr\}. \]
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Here is a proof that the polynomials $1, x, x^2$ are linearly independent in the vector space ${\mathbb P}_2$.
Assume that $\alpha_1,$ $\alpha_2,$ and $\alpha_3$ are scalars in $\mathbb{R}$ such that \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1\cdot 1 + \alpha_2 x + \alpha_3 x^2 =0 \quad \text{for all} \quad x \in \mathbb{R}}. \] The objective here is to prove \[ \bbox[5px, #FF4444, border: 1pt solid red]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0}. \] Consider the left-hand side of the above green identity as a function of $x$ and take the derivative with respect to $x$. We obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_2 + 2 \alpha_3 x =0 \quad \text{for all} \quad x \in \mathbb{R}}. \] Again, consider the left-hand side of the above green identity as a function of $x$ and take the derivative with respect to $x$. We obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{2 \alpha_3 =0 \quad \text{for all} \quad x \in \mathbb{R}}. \] Substituting $x=0$ in the first two green identities and dividing the third green equality by $2$ we obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0}. \] In this way we have greenifyed the red statement. That is, we proved it.
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Here is an alternative proof that the polynomials $1, x, x^2$ are linearly independent in the vector space ${\mathbb P}_2$.
Assume that $\alpha_1,$ $\alpha_2,$ and $\alpha_3$ are scalars in $\mathbb{R}$ such that \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1\cdot 1 + \alpha_2 x + \alpha_3 x^2 =0 \quad \text{for all} \quad x \in \mathbb{R}}. \] The objective here is to prove \[ \bbox[5px, #FF4444, border: 1pt solid red]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0}. \] The above green identity holds for all $x\in\mathbb{R}.$ In particular it holds for specific $x=-1,$ $x=0,$ and $x=1.$ That is, we have \[ \bbox[5px, #88FF88, border: 1pt solid green]{ \begin{array}{lr} \alpha_1 - \alpha_2 +\alpha_3 &=0 \\ \alpha_1 &=0 \\ \alpha_1 + \alpha_2 +\alpha_3 &=0 \\ \end{array} } \] The last green box contains a homogeneous system of linear equations which can be written in a matrix form as \[ \bbox[5px, #88FF88, border: 1pt solid green]{ \left[\!\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}\!\right] \left[\!\begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array}\!\right] = \left[\!\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\!\right] } \] Since the determinant of the above $3\!\times\!3$ matrix is $2$, the above homogeneous equation has only the trivial solution. That is, \[ \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0}. \] In this way we have greenifyed the red statement. That is, we proved it.
- Hence, \[ \mathbb{P}_2 = \operatorname{Span}\bigl\{ 1, x, x^2 \bigr\} \] and $\mathcal{M} = \bigl\{ 1, x, x^2 \bigr\}$ is a basis for $\mathbb{P}_2.$ I denoted this basis by $\mathcal{M}$ since the polynomials $1,$ $x,$ $x^2$ are called monomials.
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Let us now look at few subspaces of $\mathbb{P}_2.$ First consider the set
\[
\mathcal{H}_0 = \bigl\{ \mathbf{p} \in \mathbb{P}_2 \, : \, \mathbf{p}(0) = 0 \bigr\}.
\]
That is, by $\mathcal{H}_0$ we denote the set of all polynomials in $\mathbb{P}_2$ which have one root equal to $0.$ For example the polynomial $\mathbf{q}_1(x) = x$ is in $\mathcal{H}_0.$ Similarly, $\mathbf{q}_2(x) = x^2$ is also in $\mathcal{H}_0.$
- One task in this context is to use the definition of a subspace to prove that $\mathcal{H}_0$ is a subspace of $\mathbb{P}_2.$
- In three items below we will prove a more informative statement \[ \mathcal{H}_0 = \operatorname{Span}\bigl\{ x, x^2 \bigr\}. \]
- Proofs involving equality of two sets are done in two steps. We first prove that each polynomial in $\mathcal{H}_0$ must also be in $\operatorname{Span}\bigl\{ x, x^2 \bigr\},$ that is we prove that \[ \mathcal{H}_0 \subseteq \operatorname{Span}\bigl\{ x, x^2 \bigr\}. \] Then we prove that each polynomial in $\operatorname{Span}\bigl\{ x, x^2 \bigr\}$ must also be in $\mathcal{H}_0,$ that is we prove that \[ \operatorname{Span}\bigl\{ x, x^2 \bigr\} \subseteq \mathcal{H}_0. \] Together these two facts prove the equality of two sets.
- Assume that $\mathbf{p} \in \mathcal{H}_0.$ Then $\mathbf{p} \in \mathbb{P}_2$ and $\mathbf{p}(0) = 0.$ Therefore, for some $a_0, a_1, a_2 \in \mathbb{R}$ we have \[ \mathbf{p}(x) = a_0 + a_1 x + a_2 x^2 \quad \text{and} \quad \mathbf{p}(0) = a_0 + a_1 \, 0 + a_2 \, 0^2 = a_0 = 0. \] Since we proved that $a_0=0,$ we have $\mathbf{p}(x) = a_1 x + a_2 x^2,$ that is $\mathbf{p} \in \operatorname{Span}\bigl\{ x, x^2 \bigr\}.$ This proves \[ \mathcal{H}_0 \subseteq \operatorname{Span}\bigl\{ x, x^2 \bigr\}. \]
- Assume that $\mathbf{p} \in \operatorname{Span}\bigl\{ x, x^2 \bigr\}.$ Then there exist $\alpha_1, \alpha_2 \in \mathbb{R}$ such that $\mathbf{p}(x) = \alpha_1 x + \alpha_2 x^2.$ Since $\mathbf{p}(0) = \alpha_1 \, 0 + \alpha_2 \, 0^2 = 0$ we conclude that $\mathbf{p} \in \mathcal{H}_0.$ This proves \[ \operatorname{Span}\bigl\{ x, x^2 \bigr\} \subseteq \mathcal{H}_0. \] This item and the preceding one prove \[ \mathcal{H}_0 = \operatorname{Span}\bigl\{ x, x^2 \bigr\}. \]
- Above we expressed $\mathcal{H}_0$ as a span of two polynomials. By Theorem 1 each span is a subspace. Therefore, this is an alternative proof that $\mathcal{H}_0$ is a subspace. Since the set $\bigl\{ x, x^2 \bigr\}$ is linearly independent and since $\bigl\{ x, x^2 \bigr\}$ spans $\mathcal{H}_0$ we have that $\bigl\{ x, x^2 \bigr\}$ is a basis for $\mathcal{H}_0.$ Consequently \[ \dim \mathcal{H}_0 = 2. \]
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Consider the set
\[
\mathcal{H}_1 = \bigl\{ \mathbf{p} \in \mathbb{P}_2 \, : \, \mathbf{p}(1) = 0 \bigr\}.
\]
That is, by $\mathcal{H}_1$ we denote the set of all polynomials in $\mathbb{P}_2$ which have one root equal to $1.$ For example the polynomial $\mathbf{q}_1(x) = x-1$ is in $\mathcal{H}_1.$ Similarly, $\mathbf{q}_2(x) = x^2-1$ is also in $\mathcal{H}_1.$
- In two items below we will prove that \[ \mathcal{H}_1 = \operatorname{Span}\bigl\{ x -1 , x^2-1 \bigr\}. \]
- Assume that $\mathbf{p} \in \mathcal{H}_1.$ Then $\mathbf{p} \in \mathbb{P}_2$ and $\mathbf{p}(1) = 0.$ Therefore, for some $a_0, a_1, a_2 \in \mathbb{R}$ we have \[ \mathbf{p}(x) = a_0 + a_1 x + a_2 x^2 \quad \text{and} \quad \mathbf{p}(0) = a_0 + a_1 \,1 + a_2 \, 1^2 = a_0 + a_1 + a_2 = 0. \] Hence $a_0 = -a_1 - a_2$ and $\mathbf{p}$ can be expressed as \[ \mathbf{p}(x) = -a_1 - a_2 + a_1 x + a_2 x^2 = a_1 (x-1) + a_2 (x^2-1). \] That is $\mathbf{p} \in \operatorname{Span}\bigl\{ x-1, x^2-1 \bigr\}.$ This proves \[ \mathcal{H}_1 \subseteq \operatorname{Span}\bigl\{ x - 1, x^2 - 1 \bigr\}. \]
- Assume that $\mathbf{p} \in \operatorname{Span}\bigl\{ x - 1, x^2 -1 \bigr\}.$ Then there exist $\alpha_1, \alpha_2 \in \mathbb{R}$ such that \[ \mathbf{p}(x) = \alpha_1 (x-1) + \alpha_2 (x^2 - 1). \] Since $\mathbf{p}(1) = \alpha_1 \, 0 + \alpha_2 \, 0 = 0$ we conclude that $\mathbf{p} \in \mathcal{H}_1.$ This proves \[ \operatorname{Span}\bigl\{ x-1, x^2 - 1 \bigr\} \subseteq \mathcal{H}_1. \] This item and the preceding one prove \[ \mathcal{H}_1 = \operatorname{Span}\bigl\{ x - 1 , x^2 - 1 \bigr\}. \]
- Above we expressed $\mathcal{H}_1$ as a span of two polynomials. By Theorem 1 each span is a subspace. Therefore, this is an alternative proof that $\mathcal{H}_1$ is a subspace.
- Next we need to prove that $\bigl\{ x - 1, x^2 -1 \bigr\}$ is a linearly independent set. We use the definition of linear independence. Assume that $\alpha_1, \alpha_2 \in \mathbb{R}$ are such that \[ \alpha_1 (x-1) + \alpha_2 (x^2-1) = 0 \quad \text{for all} \quad x \in \mathbb{R}. \] Setting $x=-1$ yields $-2 \alpha_1 = 0.$ Thus, $\alpha_1 = 0.$ Substituting this in the assumption yields \[ \alpha_2 (x^2-1) = 0 \quad \text{for all} \quad x \in \mathbb{R}. \] Setting $x=0$ yields $-\alpha_2 = 0.$ Thus, $\alpha_2 = 0.$ Hence we proved $\alpha_1 = 0$ and $\alpha_2 = 0.$ This proves the linear independence of $\bigl\{ x - 1, x^2 -1 \bigr\}.$ Consequently $\bigl\{ x - 1, x^2 -1 \bigr\}$ is a basis for $\mathcal{H}_1$ and \[ \dim \mathcal{H}_1 = 2. \]
- We did Section 4.1 today. Recommended exercises: 1, 3, 5, 6, 7, 8, 9, 11 13, 15, 16, 17, 18, 21, 22, 23, 24.
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Here I recall the definition of a vector space as I stated it in class.
Definition. A nonempty set $\mathcal{V}$ is said to be a vector space over $\mathbb R$ if it satisfies the following ten axioms.
Axiom 1. (AE) There exists a function $+: \mathcal{V}\!\times\!\mathcal{V} \to \mathcal{V}.$That is, for each pair $(u,v) \in \mathcal{V}\!\times\!\mathcal{V}$ there exists a unique $u+v \in \mathcal{V}$ which is called the sum of $u$ and $v.$Axiom 2. (AA) For every $u, v, w \in \mathcal{V}$ we have $u+(v+w) = (u+v)+w$Axiom 3. (AC) For every $u, v \in \mathcal{V}$ we have $u+v = v+u$Axiom 4. (AZ) There exists $0 \in \mathcal{V}$ such that for every $v \in \mathcal{V}$ we have $v+0 = v$Axiom 5. (AO) For every $v \in \mathcal{V}$ there exists $-v \in \mathcal{V}$ such that $v+(-v) = 0$Axiom 6. (SE) There exists a function $\cdot: \mathbb{R}\!\times\!\mathcal{V} \to \mathcal{V}.$That is, for each real number $\alpha \in \mathbb R$ and each $v \in \mathcal{V}$ there exists a unique $\alpha v \in \mathcal{V}$ which is called the scalar product of $\alpha$ and $v.$Axiom 7. (SA) For every $\alpha, \beta \in \mathbb R$ and every $v \in \mathcal{V}$ we have $\alpha (\beta v) = (\alpha\beta) v$Axiom 8. (SD) For every $\alpha, \beta \in \mathbb R$ and every $v \in \mathcal{V}$ we have $(\alpha +\beta) v = \alpha v + \beta v$Axiom 9. (SD) For every $\alpha \in \mathbb R$ and every $u, v \in \mathcal{V}$ we have $\alpha (u + v) = \alpha u + \alpha v$Axiom 10. (S0) For every $v \in \mathcal{V}$ we have $1 v = v$Explanation of the abbreviations: AE--addition exists, AA--addition is associative, AC--addition is commutative, AZ--addition has zero, AO--addition has opposites, SE-- scaling exists, SA--scaling is associative, SD--scaling distributes over addition of real numbers, SD--scaling distributes over addition of vectors, SO--scaling with one.
- To illustrate the definition of a vector space we studied an exotic vector space: \[ \require{bbox} \mathcal{V} = \left\{ \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \, : \, x_1, x_2 \in \mathbb{R} \ \ \text{and} \ \ x_1, x_2 \gt 0 \right\} \] with the addition defined as \[ \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \bbox[yellow]{+} \bbox[yellow]{\left[\! \begin{array}{c} y_1 \\ y_2 \end{array} \!\right]} = \bbox[yellow]{\left[\! \begin{array}{c} x_1 y_1 \\ x_2 y_2 \end{array} \!\right]} \quad \text{for all} \quad \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]}, \bbox[yellow]{\left[\! \begin{array}{c} y_1 \\ y_2 \end{array} \!\right]} \in \mathcal{V} \] and scaling defined as \[ \alpha \bbox[yellow]{\phantom{*}} \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} = \bbox[yellow]{\left[\! \begin{array}{c} (x_1)^\alpha \\ (x_2)^\alpha \end{array} \!\right]} \quad \text{for all} \quad \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]}\in \mathcal{V} \quad \text{and for all} \quad \alpha \in \mathbb{R}. \] It is a nice exercise to verify that all the axioms of a vector space are satisfied. Interestingly, the zero vector in the vector space $\mathcal{V}$ is the vector $\bbox[yellow]{\left[\! \begin{array}{c} 1 \\ 1 \end{array} \!\right]}.$
- I colored the vectors in $\mathcal{V}$ yellow to distinguish them from the vectors in $\mathbb{R}^2.$
- Now consider the function $L:\mathcal{V} \to \mathbb{R}^2$ defined as follows \[ L \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} = \left[\! \begin{array}{c} \ln x_1 \\ \ln x_2 \end{array} \!\right] \quad \text{for all} \quad \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]}\in \mathcal{V}. \] This function has the following two properties: \[ L \left(\bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \bbox[yellow]{+} \bbox[yellow]{\left[\! \begin{array}{c} y_1 \\ y_2 \end{array} \!\right]} \right) = L \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} + L \bbox[yellow]{\left[\! \begin{array}{c} y_1 \\ y_2 \end{array} \!\right]} \quad \text{for all} \quad \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]}, \bbox[yellow]{\left[\! \begin{array}{c} y_1 \\ y_2 \end{array} \!\right]} \in \mathcal{V} \] and \[ L \left(\alpha\bbox[yellow]{\phantom{*}} \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \right) = \alpha L \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \quad \text{for all} \quad \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \in \mathcal{V} \quad \text{and for all} \quad \alpha \in \mathbb{R}. \]
- These two properties make $L:\mathcal{V} \to \mathbb{R}^2$ a linear transformation defined on $\mathcal{V}$ to $\mathbb{R}^2.$
- It is interesting to study subspaces of the vector space $\mathcal{V}.$ For example, what is \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 1 \\ 2 \end{array} \!\right]} \right\}, \] or, what is \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 2 \\ 1 \end{array} \!\right]} \right\}, \] or, \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 2 \\ 2 \end{array} \!\right]} \right\}, \] or \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 4 \\ 2 \end{array} \!\right]} \right\}, \] or, \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 2 \\ 4 \end{array} \!\right]} \right\}. \]
- I am posting two challenging problems related to this section.
- Today we did a remarkable application of eigenvalues and eigenvectors to a famous sequence of positive integers: Fibonacci numbers.
- Recall that $\mathbb{N}$ denotes the set of positive integers. The Fibonacci numbers are the elements of the sequence \[ f_0, f_1, f_2, \ldots, f_n, \ldots \] recursively defined by \[ f_0 = 0, \quad f_1 = 1, \quad \text{and} \quad f_{n+1} = f_n + f_{n-1} \quad \text{for} \quad n \in \mathbb{N}. \] Since we are given $f_0 = 0$ and $f_1 = 1$ using the recursion $f_{n+1} = f_n + f_{n-1}$ with $n=1$ we get $f_2 = 0+1 = 1.$ A repeated use of the recursion $f_{n+1} = f_n + f_{n-1}$ with $n=2$, then $n=3$, and so on, we get \[ f_0 = 0, \ f_1 = 1, \ f_2 = 1, \ f_3 = 2, \ f_4 = 3, \ f_5 = 5, \ f_6 = 8, \ f_7 = 13, \ f_8 = 21, \ f_9 = 34, \ f_{10} = 55, \ f_{11} = 89, \ \ldots \ \] It is clear that with enough patience we can calculate $f_{100}$ by calculating all Fibonacci numbers preceding it. Computers are really good with doing recursively defined operations. I wrote a webpage on how to do this in Mathematica. If you want to try using Mathematica on WWU computers please let me know. I can help you with that. By the way, \[ f_{100} = 354,224,848,179,261,915,075 \]
- Since in Mathematics we like to be able to approach mathematical concepts in diverse ways, we are interested in finding a formula for the $n$-th Fibonacci number without calculating all the preceding Fibonacci numbers; a formula that will use only $n$ and algebraic operations. Amazingly, Linear Algebra offers a way to do that. The next items will illustrate how that comes about.
- The first step is to write the recutsion \[ f_{n+1} = f_n + f_{n-1} \] using a matrix: \[ \left[\!\begin{array}{l} f_{n} \\ f_{n+1} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} f_{n-1} \\ f_{n} \end{array}\!\right]. \] Thus, we can obtain the Fibonacci sequence by repeated application of the matrix $\displaystyle\left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]$ as follows \begin{align*} \left[\!\begin{array}{l} f_{1} \\ f_{2} \end{array}\!\right] & = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} f_0 \\ f_{1} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] \\ \left[\!\begin{array}{l} f_{2} \\ f_{3} \end{array}\!\right] & = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} f_{1} \\ f_{2} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^2 \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] \\ \left[\!\begin{array}{l} f_{3} \\ f_{4} \end{array}\!\right] &= \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} f_{2} \\ f_{3} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]\left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^2 \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^3 \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] \\ \left[\!\begin{array}{l} f_{4} \\ f_{5} \end{array}\!\right] &= \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} f_{3} \\ f_{4} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^3 \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^4 \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] \\ & \ \vdots \\ \left[\!\begin{array}{l} f_{n} \\ f_{n-1} \end{array}\!\right] &= \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} f_{n-1} \\ f_{n-2} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^{n-1} \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] \\ \left[\!\begin{array}{l} f_{n+1} \\ f_{n} \end{array}\!\right] &= \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{l} f_{n} \\ f_{n-1} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^{n} \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^{n+1} \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] \end{align*}
- In the preceding item we saw that \[ \left[\!\begin{array}{l} f_{n} \\ f_{n-1} \end{array}\!\right] = \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right]. \] Therefore \[ f_n = \bigl[ 1 \ \ 0 \bigr] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right]. \] We could stop here and pronounce that this is sufficiently good formula for $f_n$ which uses only $n$ and matrix algebra. However, we want a formula for $f_n$ which uses only algebra with specific numbers; without matrices. To obtain such a formula we will calculate the eigenvalues and eigenvectors of the matrix $\displaystyle\left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]$.
- First we calculate the characteristic polynomial of $\displaystyle\left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]$: \[ \det\left( \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right] - \left[\!\begin{array}{cc} \lambda & 0 \\ 0 & \lambda \end{array}\!\right] \right) = \left|\!\begin{array}{cc} -\lambda & 1 \\ 1 & 1-\lambda \end{array}\!\right| = -\lambda(1-\lambda) - 1 = \lambda^2 - \lambda - 1 \] The roots of the characteristic polynomial are \[ \lambda_1 = \frac{1+\sqrt{5}}{2} = \varphi \quad \text{and} \quad \lambda_2 = \frac{1-\sqrt{5}}{2} = \psi. \] The number $\phi$ is the famous number Golden ratio. The Greek letter $\varphi$ is the standard notation for the Golden ratio. We introduce $\psi$ since we will use it several times below.
- An eigenvector of $\displaystyle\left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]$ corresponding to $\varphi$ is $\displaystyle\left[\!\begin{array}{l} 1 \\ \varphi \end{array}\!\right]$ and an eigenvector corresponding to $\psi$ is $\displaystyle\left[\!\begin{array}{l} 1 \\ \psi \end{array}\!\right]$. Please verify that \[ \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]\left[\!\begin{array}{l} 1 \\ \varphi \end{array}\!\right] = \varphi \left[\!\begin{array}{l} 1 \\ \varphi \end{array}\!\right] \quad \text{and} \quad \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]\left[\!\begin{array}{l} 1 \\ \psi \end{array}\!\right] = \psi \left[\!\begin{array}{l} 1 \\ \psi \end{array}\!\right]. \] To verify the preceding vector equalities you will use that $\varphi$ and $\psi$ are the roots of the characteristic polynomial, that is \[ \varphi^2 - \varphi - 1 = 0 \quad \text{and} \quad \psi^2 - \psi - 1 = 0. \] One of the important properties of eigenvectors is that it is easy to calculate the action of the powers of the matrix on eigenvectors: \[ \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{c} 1 \\ \varphi \end{array}\!\right] = \varphi^n \left[\!\begin{array}{c} 1 \\ \varphi \end{array}\!\right] \quad \text{and} \quad \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{c} 1 \\ \psi \end{array}\!\right] = \psi^n \left[\!\begin{array}{c} 1 \\ \psi \end{array}\!\right]. \]
- To improve the formula \[ f_n = \bigl[ 1 \ \ 0 \bigr] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{c} 0 \\ 1 \end{array}\!\right] \] we will represent the vector $\displaystyle\left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right]$ as a linear combination of the eigenvectors: \[ \left[\!\begin{array}{c} 0 \\ 1 \end{array}\!\right] = x_1 \left[\!\begin{array}{c} 1 \\ \varphi \end{array}\!\right] + x_2 \left[\!\begin{array}{c} 1 \\ \psi \end{array}\!\right]. \] We do not need to do row reduction to solve this system. Since $x_1+x_2 = 0$ we deduce that $x_2 = -x_1$. Then we have \[ x_1 (\varphi - \psi) = 1. \] Since $\varphi - \psi = \sqrt{5}$ we have \[ \left[\!\begin{array}{c} 0 \\ 1 \end{array}\!\right] = \frac{1}{\sqrt{5}} \left[\!\begin{array}{c} 1 \\ \varphi \end{array}\!\right] - \frac{1}{\sqrt{5}} \left[\!\begin{array}{c} 1 \\ \psi \end{array}\!\right]. \] Therefore, \begin{align*} f_n & = \bigl[ 1 \ \ 0 \bigr] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{l} 0 \\ 1 \end{array}\!\right] \\ & = \bigl[ 1 \ \ 0 \bigr] \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left( \frac{1}{\sqrt{5}} \left[\!\begin{array}{c} 1 \\ \varphi \end{array}\!\right] - \frac{1}{\sqrt{5}} \left[\!\begin{array}{c} 1 \\ \psi \end{array}\!\right] \right) \\ & = \bigl[ 1 \ \ 0 \bigr] \left( \frac{1}{\sqrt{5}} \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{c} 1 \\ \varphi \end{array}\!\right] - \frac{1}{\sqrt{5}} \left[\!\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\!\right]^n \left[\!\begin{array}{c} 1 \\ \psi \end{array}\!\right] \right) \\ & = \bigl[ 1 \ \ 0 \bigr] \left( \frac{1}{\sqrt{5}} \varphi^n \left[\!\begin{array}{l} 1 \\ \varphi \end{array}\!\right] - \frac{1}{\sqrt{5}} \psi^n \left[\!\begin{array}{c} 1 \\ \psi \end{array}\!\right] \right) \\ & = \frac{1}{\sqrt{5}} \bigl[ 1 \ \ 0 \bigr] \left( \left[\!\begin{array}{c} \varphi^n \\ \varphi^{n+1} \end{array}\!\right] - \left[\!\begin{array}{c} \psi^{n} \\ \psi^{n+1} \end{array}\!\right] \right) \\ & = \frac{1}{\sqrt{5}} \bigl[ 1 \ \ 0 \bigr] \left[\!\begin{array}{c} \varphi^n - \psi^{n} \\ \varphi^{n+1} -\psi^{n+1} \end{array}\!\right]\\ & = \frac{1}{\sqrt{5}} \bigl( \varphi^n - \psi^{n} \bigr). \end{align*} Thus, finally and amazingly, \[ f_n = \frac{1}{\sqrt{5}} \bigl( \varphi^n - \psi^{n} \bigr) = \frac{1}{\sqrt{5}} \left( \frac{(1+\sqrt{5})^n}{2^n} - \frac{(1-\sqrt{5})^n}{2^n} \right) = \frac{ (1+\sqrt{5})^n - (1-\sqrt{5})^n }{2^n \sqrt{5}} \] for all nonnegative integers $n$. A formula like this in which $f_n$ is given in terms of $n$ and standard functions is called a closed form expression. A lot of effort in mathematics has been put in finding closed form expressions for different mathematical objects.
- This and the following item are related to the Fibonacci numbers but not to Linear ALgebra. I post it since it is interesting mathematics that you might understand. Without knowing what we derived above, if somebody showed you the expression \[ \frac{ (1+\sqrt{5})^n - (1-\sqrt{5})^n }{2^n \sqrt{5}} \] with $n$ a positive integer, how would one know that this is an integer. For small $n$ we could calculate, and we would get the Fibonacci numbers, but proving that it is always the case is an interesting exercise.
- One can use the Binomial theorem to expend the numerator \[ (1+\sqrt5)^n-(1-\sqrt5)^n=\sum_{k=0}^{n} \binom{n}{k}(\sqrt{5})^k-\sum_{k=0}^{n} \binom{n}{k}(-1)^k (\sqrt{5})^k. \] In the above sums we can write the even and odd terms separately and get \[ \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j}(\sqrt{5})^{2j} + \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1}(\sqrt{5})^{2j+1} - \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j}(\sqrt{5})^{2j} + \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1}(\sqrt{5})^{2j+1}. \] Hence \[ (1+\sqrt5)^n-(1-\sqrt5)^n=2 \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1}(\sqrt{5})^{2j+1} = 2 \sqrt{5} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j}. \] Dividing by $2^n\sqrt{5}$ so we get: \[ \frac{ (1+\sqrt{5})^n - (1-\sqrt{5})^n }{2^n \sqrt{5}} = \frac{2\sqrt{5}}{2^n \sqrt{5}}\sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j} = \frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j}. \] This gives us another interesting formula for the Fibonacci numbers \[ f_n = \frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j} \quad \text{where} \quad n \in\mathbb{N}. \] But it would be interesting to prove that the number \[ \frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j} \quad \text{where} \quad n \in\mathbb{N} \] is always an integer. I do not know how to do that directly, without all the work done in this post.
- Recommended exercises for Section 5.3 are 2, 3, 5, 8, 9, 12, 13, 16, 18, 20, 23, 24.
-
Today we proved the following theorem
Theorem. Let $n \in \mathbb{N}$ and let $A$ be an $n\!\times\!n$ matrix. The matrix $A$ is diagonalizable if and only if there exists a basis of $\mathbb{R}^n$ which consists of eigenvectors of $A.$
Theorem. Let $n \in \mathbb{N}$ and let $A$ be an $n\!\times\!n$ matrix. The following two statements are equivalent:
(a) There exist an invertible $n\!\times\!n$ matrix $P$ and a diagonal $n\!\times\!n$ matrix $D$ such that $A= PDP^{-1}.$
(b) There exist linearly independent vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ in $\mathbb{R}^n$ and real numbers $\lambda_1, \lambda_2,\ldots,\lambda_n$ such that $A \mathbf{v}_k = \lambda_k \mathbf{v}_k$ for all $k\in \{1,\ldots,n\}.$
- Proof of (b)⇒(a). Assume (b). That is, assume that there exist linearly independent vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ in $\mathbb{R}^n$ and real numbers $\lambda_1, \lambda_2,\ldots,\lambda_n$ such that \[ A \mathbf{v}_k = \lambda_k \mathbf{v}_k \quad \text{for all} \quad k\in \{1,\ldots,n\}. \] The preceding displayed expression contains $n$ vector equations. These $n$ vector equations can be expressed as one matrix equation \[ \Bigl[ A \mathbf{v}_1 \ A \mathbf{v}_2 \ \cdots \ A \mathbf{v}_n \Bigr] = \Bigl[ \lambda_1 \mathbf{v}_1 \ \lambda_2 \mathbf{v}_2 \ \cdots \ \lambda_n \mathbf{v}_n \Bigr]. \] By the definition of the matrix multiplication the preceding matrix equality can be expressed as \[ A \Bigl[ \mathbf{v}_1 \ \mathbf{v}_2 \ \cdots \ \mathbf{v}_n \Bigr] = \Bigl[ \mathbf{v}_1 \ \mathbf{v}_2 \ \cdots \ \mathbf{v}_n \Bigr] \left[\!\begin{array}{cccc} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{array}\!\right]. \] The preceding matrix equality can be written as \[ A P = P D, \] where we set \[ P = \Bigl[ \mathbf{v}_1 \ \mathbf{v}_2 \ \cdots \ \mathbf{v}_n \Bigr] \quad \text{and} \quad D = \left[\!\begin{array}{cccc} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{array}\!\right]. \] Since we assume that $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ are linearly independent vectors in $\mathbb{R}^n$ the matrix $P$ is invertible by the Invertible Matrix Theorem. Therefore, the matrix $P^{-1}$ exists. Multiplying the equality \[ AP=PD \] from the right by $P^{-1}$ we get \[ A = PDP^{-1}. \] This proves (a).
- Now we recall Example 1 from Monday, March 1, 2021. We calculated that \begin{align*} \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right] & = 1 \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right], \\ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right] & = 2 \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right], \\ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right]\left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right] & = 2 \left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right]. \end{align*} and \[ \left[\! \begin{array}{crc|ccc} 1 & -1 & 1 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 1 & 0 \\ 3 & 0 & 1 & 0 & 0 & 1 \end{array} \!\right] \sim \cdots \sim \left[\! \begin{array}{ccc|rrc} 1 & 0 & 0 & -1 & -1 & 1\\ 0 & 1 & 0 & 1 & 2 & -1 \\ 0 & 0 & 1 & 3 & 3 & -2 \end{array} \!\right]. \]
- In the preceding item we have that \[ A = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \] is a $3\!\times\!3$ matrix. The vectors \[ \mathbf{v}_1 = \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right], \quad \mathbf{v}_2 = \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right], \quad \mathbf{v}_3 = \left[\! \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \!\right] \] are linearly independent vectors in $\mathbb{R}^3$ which are eigenvectors of $A$ such that \[ A \mathbf{v}_1 = 1 \mathbf{v}_1, \quad A \mathbf{v}_2 = 2 \mathbf{v}_2, \quad A \mathbf{v}_3 = 2 \mathbf{v}_3. \] Based on the proof presented above the preceding three vector equations can be written as one matrix equation as follows (this is the equation $AP=PD$ in the proof) \[ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \left[\! \begin{array}{crc} 1 & -1 & 1\\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{array} \!\right] = \left[\! \begin{array}{crc} 1 & -1 & 1\\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{array} \!\right] \left[\! \begin{array}{crc} 1 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \!\right] \] Multiplying the preceding matrix equation with the inverse calculated in the previous item \[ \left[\! \begin{array}{crc} 1 & -1 & 1\\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{array} \!\right]^{-1} = \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right], \] we get the diagonalization of the matrix $A$: \[ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] = \left[\! \begin{array}{rrr} 1 & -1 & 1 \\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{array} \!\right] \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] . \] Notice that this is a specific numerical matrix equality and it takes just a little bit of patience to verify whether it is true or not.
- For Section 5.2 do 1-8, 11, 12, 14, 15, (in all these problems you can find eigenvectors as well) 9, 13, 18, 19, 20, 21, 24, 25, 27.
-
Example 1. In this item I will illustrate how to calculate eigenvalues and the corresponding eigenspaces of a specific $3\!\times\!3$ matrix. Consider the matrix
\[
A = \left[\!
\begin{array}{rrr}
3 & 1 & -1 \\
1 & 3 & -1 \\
3 & 3 & -1
\end{array}
\!\right] .
\]
- First we find the characteristic polynomial of this matrix. The characteristic polynomial is the determinant of the following matrix: \[ A - \lambda I_3 = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] - \left[\! \begin{array}{rrr} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{array} \!\right] = \left[\! \begin{array}{ccc} 3-\lambda & 1 & -1 \\ 1 & 3-\lambda & -1 \\ 3 & 3 & -1-\lambda \end{array} \!\right] \] Next we calculate this determinant: \begin{align*} \left|\! \begin{array}{ccc} 3-\lambda & 1 & -1 \\ 1 & 3-\lambda & -1 \\ 3 & 3 & -1-\lambda \end{array} \!\right| &= \left|\! \begin{array}{ccc} 2-\lambda & -2+\lambda & 0 \\ 1 & 3-\lambda & -1 \\ 3 & 3 & -1-\lambda \end{array} \!\right| \\ &= \left|\! \begin{array}{ccc} 2-\lambda & 0 & 0 \\ 1 & 4-\lambda & -1 \\ 3 & 6 & -1-\lambda \end{array} \!\right| \\ &= (2-\lambda) \bigl( (4-\lambda)(-1-\lambda) + 6 \bigr) \\ & = (2-\lambda)\bigl(\lambda^2 - 3 \lambda + 2\bigr) \\ & = -(\lambda - 2)^2 ( \lambda - 1) \end{align*} (At the first equality sign, we subtracted the second row from the first. At the second equality sign, we added the first column to the second. These operations do not change the value of a determinant.)
- Thus the eigenvalues of the matrix $A$ are $1$ and $2.$
- Next we will find the eigenspace corresponding to the eigenvalue $1.$ For that we need to find the nullspace of the matrix \[ A - 1 I_3 = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] - \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right] = \left[\! \begin{array}{ccc} 2 & 1 & -1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right]. \] So, we row reduce this matrix: \[ \left[\! \begin{array}{ccc} 2 & 1 & -1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] \sim \left[\! \begin{array}{ccc} 1 & 2 & -1 \\ 0 & 3 & -1 \\ 0 & 3 & -1 \end{array} \!\right] \sim \left[\! \begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{array} \!\right] \sim \left[\! \begin{array}{ccc} 1 & 0 & -1/3 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{array} \!\right]. \] Thus, the eigenspace is the subspace \[ \left\{ \left[\! \begin{array}{c} s/3 \\ s/3 \\ s \end{array} \!\right] \ : \ s \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right] \right\}. \] Hence one eigenvector is $\left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right].$
- Next we will find the eigenspace corresponding to the eigenvalue $2.$ For that we need to find the nullspace of the matrix \[ A - 2 I_3 = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] - \left[\! \begin{array}{rrr} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \!\right] = \left[\! \begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 3 & 3 & -3 \end{array} \!\right]. \] So, we row reduce this matrix: \[ \left[\! \begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 3 & 3 & -3 \end{array} \!\right] \sim \left[\! \begin{array}{rrr} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \!\right]. \] Thus, the eigenspace is the subspace \[ \left\{ \left[\! \begin{array}{c} -s + t \\ s \\ t \end{array} \!\right] \ : \ s, t \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \!\right], \left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right] \right\}. \] Hence two linearly independent eigenvectors corresponding to the eigenvalue $2$ are $\left[\! \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \!\right]$ and $\left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right].$
- The magic of what we found by now is that we found a basis of $\mathbb{R}^3$ which consists of eigenvectors of $A:$ \[ \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right], \quad \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right], \quad \left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right]. \]
- It is easy to verify whether these are really eigenvectors: \begin{align*} \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right] & = 1 \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right], \\ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right] & = 2 \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right], \\ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right]\left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right] & = 2 \left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right]. \end{align*} Yes, they are.
- I also made the claim that the three eigenvectors are linearly independent. Let us verify that as well. \[ \left[\! \begin{array}{crc|ccc} 1 & -1 & 1 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 1 & 0 \\ 3 & 0 & 1 & 0 & 0 & 1 \end{array} \!\right] \sim \cdots \sim \left[\! \begin{array}{ccc|rrc} 1 & 0 & 0 & -1 & -1 & 1\\ 0 & 1 & 0 & 1 & 2 & -1 \\ 0 & 0 & 1 & 3 & 3 & -2 \end{array} \!\right]. \] We know that the right-hand side matrix in the Reduced Row Echelon Form is the inverse of the matrix whose columns are the eigenvectors. To verify the row reduction above, we calculate: \[ \left[\! \begin{array}{rrr} 1 & -1 & 1 \\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] = \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right]. \]
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Example 2. In this item I will illustrate how to calculate eigenvalues and the corresponding eigenspaces of a specific $4\!\times\!4$ matrix. The purpose is to demonstrate a matrix that is not diagonalizable. Consider the matrix
\[
A = \left[\!
\begin{array}{rrrr}
0 & 0 & -1 & -1 \\
-1 & 0 & 0 & 0 \\
2 & 1 & 2 & 1 \\
-2 & -1 & -1 & 0
\end{array}
\!\right] .
\]
- First we find the characteristic polynomial of this matrix. The characteristic polynomial is the determinant of the following matrix: \[ A - \lambda I_4 = \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] - \left[\! \begin{array}{rrrr} \lambda & 0 &0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{array} \!\right] = \left[\! \begin{array}{cccc} -\lambda & 0 & -1 & -1 \\ -1 & -\lambda & 0 & 0 \\ 2 & 1 & 2-\lambda & 1 \\ -2 & -1 & -1 & -\lambda \end{array} \!\right] \] Next we calculate the determinant of the preceding matrix: \begin{align*} \left|\! \begin{array}{cccc} -\lambda & 0 & -1 & -1 \\ -1 & -\lambda & 0 & 0 \\ 2 & 1 & 2-\lambda & 1 \\ -2 & -1 & -1 & -\lambda \end{array} \!\right| & = -\lambda \left|\begin{array}{ccc} -\lambda & 0 & 0 \\ 1 & 2-\lambda & 1 \\ -1 & -1 & -\lambda \end{array} \right| - \left| \begin{array}{ccc} -1 & -\lambda & 0 \\ 2 & 1 & 1 \\ -2 & -1 & -\lambda \end{array} \right| + \left| \begin{array}{ccc} -1 & -\lambda & 0 \\ 2 & 1 & 2-\lambda \\ -2 & -1 & -1 \end{array} \right| \\ & = \lambda^2 ( \lambda^2 -2 \lambda + 1 ) - \bigl( \lambda -1 +2 \lambda - 2 \lambda^2 \bigr) + \bigl( \lambda - 1 + 2 \lambda - 2 \lambda^2 \bigr) \\ & = \lambda^2 ( \lambda^2 -2 \lambda + 1 ) \\ & = \lambda^2 ( \lambda - 1 )^2 \end{align*}
- Thus the eigenvalues of the matrix $A$ are $0$ and $1.$
- Next we will find the eigenspace corresponding to the eigenvalue $0.$ For that we need to find the nullspace of the matrix \[ A - 0 I_4 = \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] \]
- So, we row reduce the matrix $A:$ \[ \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] \sim \cdots \sim \left[\! \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \!\right] \] Thus, the eigenspace is the subspace \[ \left\{ \left[\! \begin{array}{c} 0 \\ s \\ -s \\ s \end{array} \!\right] \ : \ s \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{r} 0 \\ 1 \\ -1 \\ 1 \end{array} \!\right] \right\}. \]
- Thus, eigenspace corresponding to the eigenvalue $0$ is one-dimensional.
- Next we will find the eigenspace corresponding to the eigenvalue $1.$ For that we need to find the nullspace of the matrix \[ A - 1 I_4 = \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] - \left[\! \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \!\right] = \left[\! \begin{array}{rrrr} -1 & 0 & -1 & -1 \\ -1 & -1 & 0 & 0 \\ 2 & 1 & 1 & 1 \\ -2 & -1 & -1 & -1 \end{array} \!\right] \]
- So, we row reduce the last matrix: \[ \left[\! \begin{array}{rrrr} -1 & 0 & -1 & -1 \\ -1 & -1 & 0 & 0 \\ 2 & 1 & 1 & 1 \\ -2 & -1 & -1 & -1 \end{array} \!\right] \sim \cdots \sim \left[\! \begin{array}{rrrr} 1 & 0 & 1 & 1 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \!\right] \] Thus, the eigenspace corresponding to the eigenvalue $1$ is the subspace \[ \left\{ \left[\! \begin{array}{c} -s-t \\ s+t \\ s \\ t \end{array} \!\right] \ : \ s, t \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{r} -1 \\ 1 \\ 1 \\ 0 \end{array} \!\right], \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \\ 1 \end{array} \!\right] \right\}. \]
- Thus, the eigenspace corresponding to the eigenvalue $1$ is two-dimensional.
- Since we found all eigenspaces of the $4\!\times\!4$ matrix $A$ and these eigenspaces have dimensions $1$ and $2$, we conclude that we can have at most three linearly independent eigenvectors. Consequently, we can not have a basis for $\mathbb R^4$ which consists of eigenvectors of $A.$ This shows directly that the matrix $A$ is not diagonalizable.
- Read Section 5.1. Suggested problems for Section 5.1: 1, 3, 4, 5, 6, 8, 11, 15, 16, 17, 19, 20, 24-27, 29, 30, 31.
- A related Wikipedia link: Eigenvalue, eigenvector and eigenspace.
- Below are animations of different matrices in action. In each scene the navy blue vector is the image of the sea green vector under the multiplication by a matrix $A$. For easier visualization of the action the heads of vectors leave traces.
- Just looking at the movies you can guess what are the eigenvalues and eigenvectors of the featured matrix. In particular it is easy to see whether an eigenvalue is positive, negative, zero, or complex, ... You can also approximately calculate which matrix is featured in each movie.
Place the cursor over the image to start the animation.
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Related to one of the questions on Assignment 2, today we discussed a general strategy for a problem as follows:
Let \[ \mathbf{a}_1 = \left[\!\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\!\right], \ \ \mathbf{a}_2 = \left[\!\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\!\right], \quad \mathbf{b}_1 = \left[\!\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\!\right], \ \ \mathbf{b}_2 = \left[\!\begin{array}{c} 3 \\ 2 \\ 1 \end{array}\!\right] \in \mathbb{R}^3, \] Let \[ \mathcal{G} = \operatorname{Span}\bigl\{ \mathbf{a}_1,\mathbf{a}_2 \bigr\} \quad \text{and} \quad \mathcal{H} = \operatorname{Span}\bigl\{ \mathbf{b}_1,\mathbf{b}_2 \bigr\}. \] Consider the subspace \[ \mathcal{G} \cap \mathcal{H} = \bigl\{ \mathbf{v} \in \mathbb{R}^n \ : \ \mathbf{v} \in \mathcal{G} \ \ \text{and} \ \ \mathbf{v} \in \mathcal{H} \bigr\}. \] Find a basis for $\mathcal{G} \cap \mathcal{H}.$
- We seek $\mathbf{v} \in \mathbb{R}^n$ such that there exist \[ \alpha_1, \alpha_2, \beta_1, \beta_2, \in \mathbb{R} \] such that \[ \mathbf{v} = \alpha_1 \mathbf{a}_1 + \alpha_2 \mathbf{a}_2 \quad \text{and} \quad \mathbf{v} = \beta_1 \mathbf{b}_1 + \beta_2 \mathbf{b}_2. \] But, then \[ \mathbf{0} = \mathbf{v} - \mathbf{v} = \alpha_1 \mathbf{a}_1 + \alpha_2 \mathbf{a}_2 + (-\beta_1) \mathbf{b}_1 + (-\beta_2) \mathbf{b}_2. \] Therefore, what we are seeking is related to the homogeneous vector equation \[ x_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 + x_3 \mathbf{b}_1 + x_4 \mathbf{b}_2 = \mathbf{0}. \] We know how to solve this equation by row reducing the matrix \[ \require{bbox} \left[\!\begin{array}{cccc} 1 & 0 & 1 & 3 \\ 1 & 1 & 2 & 2 \\ 0 & 1 & 3 & 1\end{array}\!\right] \sim \cdots \sim \left[\!\begin{array}{cccr|c} 1 & 0 & 0 & \bbox[lightblue]{2} \\ 0 & 1 & 0 & \bbox[lightblue]{-2} \\ 0 & 0 & 1 & \bbox[lightblue]{1} \end{array}\!\right]. \] What does the preceding Reduced Row Echelon Form tells us about the linear dependence of the vectors \[ \mathbf{a}_1 = \left[\!\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\!\right], \ \ \mathbf{a}_2 = \left[\!\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\!\right], \quad \mathbf{b}_1 = \left[\!\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\!\right], \ \ \mathbf{b}_2 = \left[\!\begin{array}{c} 3 \\ 2 \\ 1 \end{array}\!\right] ? \] I hope that discovering this linear dependence you will recognize one vector in $\mathcal{G} \cap \mathcal{H}.$ Then you should be able to make an argument that that vector spans $\mathcal{G} \cap \mathcal{H}.$
- I am fascinated by the pattern that I will present in the next two items. The pattern that I present is based on, or justifies the definition of the determinant.
- Let \[ A = \left[\! \begin{array}{cc} a & b \\ c & d \end{array} \!\right]. \] Then without any conditions on the matrix $A$ we have \[ \left[\! \begin{array}{cc} a & b \\ c & d \end{array} \!\right] \left[\! \begin{array}{rr} d & -b \\ -c & a \end{array} \!\right] = (ad-bc) \left[\! \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \!\right]. \] Therefore $A$ is invertable if and only if $\det(A) = ad-bc \neq 0.$ If $\det(A) = ad-bc \neq 0,$ then \[ A^{-1} = \frac{1}{ad-bc} \left[\! \begin{array}{rr} d & -b \\ - c & a \end{array} \!\right] \]
- It is truly amazing that the same pattern holds for a $3\!\times\!3$ matrix. We just have to take into account that the definition of the determinant for a $3\!\times\!3$ matrix is more complicated and the determinant can be expressed as three different looking cofactor expensions.
- Let \[ A = \left[\! \begin{array}{rrr} a & b & c \\ d & e & f \\ g & h & i \end{array} \!\right]. \] Recall that \begin{align*} \require{bbox} \det(A) & = a \left(\bbox[yellow]{\left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right|}\right) + b \left(\bbox[yellow]{-\left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right|}\right) + c \left(\bbox[yellow]{\left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|}\right) \\ & = d \left(\bbox[cyan]{-\left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right|} \right) + e \left(\bbox[cyan]{\left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right|}\right) + f \left(\bbox[cyan]{-\left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|}\right) \\ & = g \left(\bbox[#FF88FF]{\left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right|}\right) + h \left(\bbox[#FF88FF]{-\left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right|}\right) + i \left(\bbox[#FF88FF]{\left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|}\right) \\ & = aei + bfg +cdh - afh - bdi - ceg. \end{align*}
- Then without any assumptions on the $3\!\times\!3$ matrix $A$ we have \[ \require{bbox} \left[\! \begin{array}{rrr} a & b & c \\ d & e & f \\ g & h & i \end{array} \!\right] \left[\! \begin{array}{rrr} \bbox[yellow]{\left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right|} & \bbox[cyan]{ -\left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right|} \\[6pt] \bbox[yellow]{ -\left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right|} & \bbox[cyan]{ \left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right|} & \bbox[#FF88FF]{ -\left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right|} \\[6pt] \bbox[yellow]{ \left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|} & \bbox[cyan]{ -\left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|} \end{array} \!\right] = \] (here we use the definition of matrix multiplication to calculate the product) \[ = \require{bbox} \left[\! \begin{array}{ccc} \bbox[yellow]{ a \left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right| - b \left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right| + c \left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|} \ & \bbox[cyan]{ - a \left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right| + b \left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right| - c \left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|} \ & \bbox[#FF88FF]{ a \left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right| - b \left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right| + c \left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|} \\[14pt] \bbox[yellow]{ d \left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right| - e \left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right| + f \left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|} \ & \bbox[cyan]{ - d \left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right| + e \left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right| - f \left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|} \ & \bbox[#FF88FF]{ d \left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right| - e \left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right| + f \left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|} \\[14pt] \bbox[yellow]{ g \left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right| - h \left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right| + i \left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|} \ & \bbox[cyan]{ - g \left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right| + h \left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right| - i \left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|} \ & \bbox[#FF88FF]{ g \left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right| - h \left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right| + i \left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|} \end{array} \!\right] = \] (now we notice that each colored entry in the preceding matrix is a cofactor expension of a $3\!\times\!3$ determinant; so each colored entry we write as a determinant; so the entry of the preceding matrix are $3\!\times\!3$ determinants) \[ = \left[\! \begin{array}{ccc} \bbox[yellow]{\left|\! \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \!\right|} & \bbox[cyan]{\left|\! \begin{array}{ccc} a & b & c \\ a & b & c \\ g & h & i \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{ccc} a & b & c \\ d & e & f \\ a & b & c \end{array} \!\right|} \\[6pt] \bbox[yellow]{ \left|\! \begin{array}{ccc} d & e & f \\ d & e & f \\ g & h & i \end{array} \!\right| } & \bbox[cyan]{\left|\! \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{ccc} a & b & c \\ d & e & f \\ d & e & f \end{array} \!\right|} \\[6pt] \bbox[yellow]{\left|\! \begin{array}{ccc} g & h & i \\ d & e & f \\ g & h & i \end{array} \!\right|} & \bbox[cyan]{ \left|\! \begin{array}{ccc} a & b & c \\ g & h & i \\ g & h & i \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \!\right| } \end{array} \!\right] = \] (now looking carefully at each of nine determinants, we see that on the diagonal we have $\det(A)$ and off-diagonal we have determinants with two identical rows, so these are equal to $0$) \[ =\left[\! \begin{array}{ccc} \bbox[yellow]{ \det(A)} & \bbox[cyan]{0} & \bbox[#FF88FF]{0} \\[6pt] \bbox[yellow]{ 0} & \bbox[cyan]{\det(A)} & \bbox[#FF88FF]{0} \\[6pt] \bbox[yellow]{0} & \bbox[cyan]{0} & \bbox[#FF88FF]{\det(A)} \end{array} \!\right] = \det(A) \left[\! \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right]. \]
- In conclusion, without any conditions on the $3\!\times\!3$ matrix $A$ we have \[ \require{bbox} \left[\! \begin{array}{rrr} a & b & c \\ d & e & f \\ g & h & i \end{array} \!\right] \left[\! \begin{array}{rrr} \bbox[yellow]{\left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right|} & \bbox[cyan]{ -\left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right|} \\[6pt] \bbox[yellow]{ -\left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right|} & \bbox[cyan]{ \left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right|} & \bbox[#FF88FF]{ -\left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right|} \\[6pt] \bbox[yellow]{ \left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|} & \bbox[cyan]{ -\left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|} \end{array} \!\right] = \det(A) \left[\! \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right]. \]
- The preceding formula shows that $A$ is invertible if and only if $\det(A) \neq 0.$ If $\det(A) \neq 0,$ then \[ A^{-1} = \frac{1}{\det(A)} \left[\! \begin{array}{rrr} \bbox[yellow]{\left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right|} & \bbox[cyan]{ -\left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right|} \\[6pt] \bbox[yellow]{ -\left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right|} & \bbox[cyan]{ \left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right|} & \bbox[#FF88FF]{ -\left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right|} \\[6pt] \bbox[yellow]{ \left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|} & \bbox[cyan]{ -\left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|} & \bbox[#FF88FF]{ \left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|} \end{array} \!\right]. \] Notice that in the matrix on the right we have cofactors of the matrix $A.$
- The preceding formula for the inverse is commonly written with the transpose of the matrix involving the cofactors. \[ A^{-1} = \frac{1}{\det(A)} \left[\! \begin{array}{rrr} \bbox[yellow]{\left|\! \begin{array}{cc} e & f \\ h & i \end{array} \!\right|} & \bbox[yellow]{-\left|\! \begin{array}{cc} d & f \\ g & i \end{array} \!\right|} & \bbox[yellow]{\left|\! \begin{array}{cc} d & e \\ g & h \end{array} \!\right|} \\ \bbox[cyan]{-\left|\! \begin{array}{cc} b & c \\ h & i \end{array} \!\right|} & \bbox[cyan]{\left|\! \begin{array}{cc} a & c \\ g & i \end{array} \!\right|} & \bbox[cyan]{-\left|\! \begin{array}{cc} a & b \\ g & h \end{array} \!\right|} \\ \bbox[#FF88FF]{\left|\! \begin{array}{cc} b & c \\ e & f \end{array} \!\right|} & \bbox[#FF88FF]{-\left|\! \begin{array}{cc} a & c \\ d & f \end{array} \!\right|} & \bbox[#FF88FF]{\left|\! \begin{array}{cc} a & b \\ d & e \end{array} \!\right|} \end{array} \!\right]^{\mathbf{\top}}. \] Please recall the definition of the cofactors of $A$. With the definition of the cofactors the above formula for the inverse is simply \[ A^{-1} = \frac{1}{\det(A)} \left[\! \begin{array}{rrr} \bbox[yellow]{C_{11}} & \bbox[yellow]{C_{12}} & \bbox[yellow]{C_{13}} \\ \bbox[cyan]{C_{21}} & \bbox[cyan]{C_{22}} & \bbox[cyan]{C_{23}} \\ \bbox[#FF88FF]{C_{31}} & \bbox[#FF88FF]{C_{32}} & \bbox[#FF88FF]{C_{33}} \end{array} \!\right]^{\mathbf{\top}}. \]
- The pattern presented above for a $3\!\times\!3$ matrix holds for any $n\!\times\!n$ matrix: \[ \left[\! \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array} \!\right] \left[\! \begin{array}{cccc} C_{11} & C_{21} & \cdots & C_{n1} \\ C_{12} & C_{22} & \cdots & C_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ C_{1n} & C_{2n} & \cdots & C_{nn} \end{array} \!\right] = \det(A) I_n. \] A justification for this identity is the definition of determinant using the cofactor expansions along the rows of $A$ and the fact that a determinant with two identical rows is equal to $0.$
- Let us calculate the inverse of a specific $3\!\times\!3$ matrix. Consider \[ A = \left[\! \begin{array}{ccc} 1 & 2 & 6 \\ 3 & 5 & 7 \\ 4 & 8 & 9 \end{array} \!\right]. \] Let us first evaluate the determinant of this matrix in three different ways, by performing the cofactor expansion along each column: \begin{align*} \det A & = 1*(\bbox[yellow]{+(-11)})+2*(\bbox[yellow]{-(-1)})+6*(\bbox[yellow]{+(4)}) \\ & = 3*(\bbox[cyan]{-(-30)})+5*(\bbox[cyan]{+(-15)})+7*(\bbox[cyan]{-(0)}) \\ & = 4*(\bbox[#FF88FF]{+(-16)})+8*(\bbox[#FF88FF]{-(-11)})+9*(\bbox[#FF88FF]{+(-1)}) \\ & = 15. \end{align*} Since we are doing this three different ways, there is a lot of checking going on in the above calculations.
- Based on the three cofactor expansions above we can write the inverse of the given matrix $A:$ \[ A^{-1} = \frac{1}{15} \left[\! \begin{array}{rrr} \bbox[yellow]{-11} & \bbox[cyan]{30} & \bbox[#FF88FF]{-16} \\ \bbox[yellow]{1} & \bbox[cyan]{-15} & \bbox[#FF88FF]{11} \\ \bbox[yellow]{4} & \bbox[cyan]{0} & \bbox[#FF88FF]{-1} \end{array} \!\right]. \] To verify, calculate \[ \left[\! \begin{array}{ccc} 1 & 2 & 6 \\ 3 & 5 & 7 \\ 4 & 8 & 9 \end{array} \!\right] \left[\! \begin{array}{rrr} \bbox[yellow]{-11} & \bbox[cyan]{30} & \bbox[#FF88FF]{-16} \\ \bbox[yellow]{1} & \bbox[cyan]{-15} & \bbox[#FF88FF]{11} \\ \bbox[yellow]{4} & \bbox[cyan]{0} & \bbox[#FF88FF]{-1} \end{array} \!\right] = \left[\! \begin{array}{rrr} \bbox[yellow]{15} & \bbox[cyan]{0} & \bbox[#FF88FF]{0} \\ \bbox[yellow]{0} & \bbox[cyan]{15} & \bbox[#FF88FF]{0} \\ \bbox[yellow]{0} & \bbox[cyan]{0} & \bbox[#FF88FF]{15} \end{array} \!\right] \]
- Suggested problems for Section 3.2: 5, 7, 9, 11, 16-20 (even), 21, 25, 31, 33, 34, 35, 40c
- Let E be an elementary matrix obtained from the identity matrix by switching two rows. The determinant of this matrix is -1. Here is a proof in four pictures.
-
Below is a "click-by-click" proof. There are nine steps in this proof. I describe each step below.
- This is the determinant that we want to calculate.
- I emphasize that the $i$-th and $j$-th row in the identity matrix are interchanged.
- We will calculate the $n\!\times\!n$ determinant by cofactor expansion along the $i$-th row.
- Since the only nonzero entry in the $i$-th row is at the $j$-th place, the cofactor expansion equals $(-1)^{i+j}$ multiplied by an $(n-1)\times(n-1)$ determinant.
- We will calculate the $(n-1)\!\times\!(n-1)$ determinant using cofactor expansion along the $(j-1)$-st row. Notice that the only nonzero entry in the $(j-1)$-st row is $1$ which is at $i$-th position.
- The previous $(n-1)\!\times\!(n-1)$ determinant calculates to $(-1)^{i+j-1}$ multiplied by the $(n-2)\!\times\!(n-2)$ determinant of the identity matrix.
- The $(n-2)\!\times\!(n-2)$ determinant of the identity matrix calculates to $1$.
- $(-1)^{i+j}(-1)^{i+j-1} = (-1)^{2i+2j-1}$.
- $(-1)^{2i+2j-1} = -1$.
All entries left blank in the determinant below are zeros.
Click on the image for a step by step proof.
- Look at Wikipedia's Matrix page. It contains some stuff that we did and much that we didn't do.
- Suggested problems for Section 3.1: 1, 3, 5, 9, 11, 13, 17, 20, 21, 22, 25-32, 37, 41, 42
-
Today in class we explored a problem inspired by the picture below. No numbers are given just the picture. In the picture below we are given two bases of $\mathbb{R}^2$, one blue and one purple:
\[
\color{blue}{\mathcal B} = \bigl\{ \color{blue}{\mathbf{b}_1}, \color{blue}{\mathbf{b}_2} \bigr\}, \quad \color{purple}{\mathcal C} = \bigl\{ \color{purple}{\mathbf{c}_1}, \color{purple}{\mathbf{c}_2} \bigr\}
\]
- Suggested exercises for Section 4.7: Change of Basis are 2, 3, 4, 6, 8, 9, 11, 12, 13, 14, 15, 16, 17, 19, 20.
- A brief review of the Change of Coordinates Matrix (this is my preferred name) follows. Let $m, n \in \mathbb{N}$ and $m\leq n$. Let $\mathcal{H}$ be a subspace of $\mathbb{R}^n$ and let \[ \mathcal{A} = \bigl\{\mathbf{a}_1,\ldots,\mathbf{a}_m\bigr\} \] and \[ \mathcal{B} = \bigl\{\mathbf{b}_1,\ldots,\mathbf{b}_m\bigr\} \] be two bases of $\mathcal{H}.$ By definition of a basis this implies that \[ \mathcal{H} = \operatorname{Span}\bigl\{\mathbf{a}_1,\ldots,\mathbf{a}_m\bigr\} = \operatorname{Span}\bigl\{\mathbf{b}_1,\ldots,\mathbf{b}_m\bigr\} \] and both \[ \mathcal{A} = \bigl\{\mathbf{a}_1,\ldots,\mathbf{a}_m\bigr\} \quad \text{and} \quad \mathcal{B} = \bigl\{\mathbf{b}_1,\ldots,\mathbf{b}_m\bigr\} \] are linearly independent. We proved in class that the change of coordinates matrix $\displaystyle\underset{\mathcal{B}\leftarrow\mathcal{A}}{P}$ is given by \[ \underset{\mathcal{B}\leftarrow\mathcal{A}}{P} = \Bigl[ \bigl[\mathbf{a}_1\bigr]_{\mathcal{B}} \ \cdots \ \bigl[ \mathbf{a}_m\bigr]_{\mathcal{B}} \Bigr] \] and analogously \[ \underset{\mathcal{A}\leftarrow\mathcal{B}}{P} = \Bigl[ \bigl[\mathbf{b}_1\bigr]_{\mathcal{A}} \ \cdots \ \bigl[ \mathbf{b}_m\bigr]_{\mathcal{A}} \Bigr]. \] But, how to calculate \[ \bigl[\mathbf{a}_1\bigr]_{\mathcal{B}}, \ldots,\bigl[ \mathbf{a}_m\bigr]_{\mathcal{B}}? \] Let us look at \[ \bigl[\mathbf{a}_1\bigr]_{\mathcal{B}} = \left[\!\begin{array}{c} x_1 \\ \vdots \\ x_m \end{array}\!\right]. \] To find the real numbers $x_1, \ldots, x_m$ we have to solve the nonhomogeneous vector equation \[ x_1 \mathbf{b}_1 + \cdots + x_m \mathbf{b}_m = \mathbf{a}_1. \] To solve the preceding equation we row reduce \[ \Bigl[\!\begin{array}{ccc|c} \mathbf{b}_1 & \cdots & \mathbf{b}_m & \mathbf{a}_1\end{array}\!\Bigr]. \] To solve the corresponding nonhomogeneous vector equations with $\mathbf{a}_2, \ldots, \mathbf{a}_m$ we just build the bigger augmented matrix: \[ \Bigl[\!\begin{array}{ccc|ccc} \mathbf{b}_1 & \cdots & \mathbf{b}_m & \mathbf{a}_1 & \cdots & \mathbf{a}_m \end{array}\!\Bigr]. \] Since the vectors $\mathbf{b}_1, \ldots, \mathbf{b}_m$ are linearly independent, the RREF off the matrix whose columns are the vectors of $\mathcal{B}$ consists of the identity matrix $I_m$ and $n-m$ zero rows at the bottom if $n\gt m.$ Therefore \[ \Bigl[\!\begin{array}{ccc|ccc} \mathbf{b}_1 & \cdots & \mathbf{b}_m & \mathbf{a}_1 & \cdots & \mathbf{a}_m \end{array}\!\Bigr] \sim \cdots \sim \left[\! \begin{array}{c|c} I_m & \underset{\mathcal{B}\leftarrow\mathcal{A}}{P} \\ 0 & 0 \end{array} \!\right]. \] The zero matrices at the bottom are present only if $n-m \gt 0$ and these matrices in this case are of size $(n-m)\!\times\!m.$
-
In the next example we are given two bases of a two-dimensional subspace of $\mathbb{R}^4$ and we are asked to find a change of coordinate matrices between these two bases:
\[
\mathcal{H} = \operatorname{Span}\left\{\left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right], \left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right]\right\} = \operatorname{Span}\left\{\left[\!\begin{array}{c}
2 \\ 5 \\ 3 \\ 7
\end{array}\!\right], \left[\!\begin{array}{r}
1 \\ 0 \\ -1 \\ 1
\end{array}\!\right]\right\}.
\]
Set
\[
\mathcal{A} = \left\{\left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right], \left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right]\right\}, \qquad \mathcal{B} = \left\{\left[\!\begin{array}{c}
2 \\ 5 \\ 3 \\ 7
\end{array}\!\right], \left[\!\begin{array}{r}
1 \\ 0 \\ -1 \\ 1
\end{array}\!\right]\right\}.
\]
To calculate $\displaystyle\underset{\mathcal{B}\leftarrow\mathcal{A}}{P}$ we need to row reduce the matrix
\[
\left[\!\begin{array}{cr|cc}
2 & 1 & 1 & 2 \\
5 & 0 & 2 & 3 \\
3 & -1 & 1 & 1 \\
7 & 1 & 3 & 5
\end{array}\!\right]
\]
The RREF of the preceding matrix will certainly include fractions. Therefore we rather find $\displaystyle\underset{\mathcal{A}\leftarrow\mathcal{B}}{P}$ for which we need to row reduce (without fractions)
\[
\left[\!\begin{array}{cc|cr}
1 & 2 & 2 & 1 \\
2 & 3 & 5 & 0 \\
1 & 1 & 3 & -1 \\
3 & 5 & 7 & 1
\end{array}\!\right] \sim \cdots \sim
\left[\!\begin{array}{cc|rr}
1 & 0 & 4 & -3 \\
0 & 1 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\!\right].
\]
Hence
\[
\underset{\mathcal{A}\leftarrow\mathcal{B}}{P} = \left[\!\begin{array}{rr}
4 & -3 \\
-1 & 2
\end{array}\!\right].
\]
Let us verify this calculation. Is it true that:
\[
\left[\!\begin{array}{c}
2 \\ 5 \\ 3 \\ 7
\end{array}\!\right] = (4) \left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right] + (-1)\left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right], \qquad \left[\!\begin{array}{r}
1 \\ 0 \\ -1 \\ 1
\end{array}\!\right] = (-3) \left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right] + (2) \left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right]?
\]
Yes. Therefore the following equalites are correct:
\[
\bigl[ \mathbf{b}_1\bigr]_{\mathcal{A}} = \left[\!\begin{array}{r}
4 \\ -1
\end{array}\!\right], \qquad \bigl[ \mathbf{b}_2\bigr]_{\mathcal{A}} = \left[\!\begin{array}{r}
-3 \\ 2
\end{array}\!\right].
\]
Hence, $\displaystyle\underset{\mathcal{A}\leftarrow\mathcal{B}}{P}$ is correct.
We have \[ \underset{\mathcal{B}\leftarrow\mathcal{A}}{P} = \left(\underset{\mathcal{A}\leftarrow\mathcal{B}}{P}\right)^{-1} = \frac{1}{5} \left[\!\begin{array}{rr} 2 & 3 \\ 1 & 4 \end{array}\!\right]. \] Verify this: \[ \left[\!\begin{array}{c} 1 \\ 2 \\ 1 \\ 3 \end{array}\!\right] = \frac{2}{5}\left[\!\begin{array}{c} 2 \\ 5 \\ 3 \\ 7 \end{array}\!\right] + \frac{1}{5} \left[\!\begin{array}{r} 1 \\ 0 \\ -1 \\ 1 \end{array}\!\right], \qquad \left[\!\begin{array}{c} 2 \\ 3 \\ 1 \\ 5 \end{array}\!\right] = \frac{3}{5} \left[\!\begin{array}{c} 2 \\ 5 \\ 3 \\ 7 \end{array}\!\right] + \frac{4}{5} \left[\!\begin{array}{r} 1 \\ 0 \\ -1 \\ 1 \end{array}\!\right]. \] True. Therefore the following equalites are correct: \[ \bigl[ \mathbf{a}_1\bigr]_{\mathcal{B}} = \frac{1}{5}\left[\!\begin{array}{r} 2 \\ 1 \end{array}\!\right], \qquad \bigl[ \mathbf{a}_2\bigr]_{\mathcal{B}} = \frac{1}{5} \left[\!\begin{array}{r} 3 \\ 4 \end{array}\!\right]. \] Hence, $\displaystyle\underset{\mathcal{B}\leftarrow\mathcal{A}}{P}$ is correct.
-
I rewrote
- In this new version I dropped one nonpivot column from the matrix $A.$ This lowered the dimension of $\operatorname{Nul}(A).$ I like it better this way. I hope I made all the corrections correctly.
-
How to make a $5\!\times\!7$ matrix in Reduced Row Echelon Form?
- First chose how many pivot columns we want. Recall that the pivot columns are the columns of the identity matrix \[ I_5 = \left[\!\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\!\right] \] The number of pivot columns must be smaller or equal than both $5$ and $7.$ Let us choose $3$ pivot columns. These must be the first three columns of the identity matrix $I_5:$ \[ \require{bbox} \left[\! \begin{array}{rrrrr} \bbox[yellow]{\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}} & \bbox[yellow]{\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}} & \bbox[yellow]{\begin{array}{r} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}} & \bbox[white]{\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{array}} & \bbox[white]{\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}} \end{array} \!\right] \]
- Now choose the positions in $5\!\times\!7$ matrix where we want these pivot columns placed. We just need to chose three numbers from $\{1,2,3,4,5,6,7\}.$ There are \[ \binom{7}{3} = \frac{7!}{3!\, 4!} = \frac{5\cdot 6 \cdot 7}{1\cdot 2\cdot 3} = 35 \] ways to choose, from $\{1,2,3\},$ $\{1,2,4\},$ $\{1,2,5\},$ $\{1,2,6\},$ $\{1,2,7\},$ $\{1,3,4\},$ all the way to $\{5,6,7\}.$ Let us choose $\{2,4,5\}$ and place the chosen columns at those positions: \[ \require{bbox} \left[\! \begin{array}{rrrrrrr} \bbox[#FF8888]{\begin{array}{c} ? \\ ? \\ ? \\ ? \\ ? \end{array}} & \bbox[yellow]{\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}} & \bbox[#FF8888]{\begin{array}{c} ? \\ ? \\ ? \\ ? \\ ? \end{array}} & \bbox[yellow]{\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}} & \bbox[yellow]{\begin{array}{r} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}} & \bbox[#FF8888]{\begin{array}{c} ? \\ ? \\ ? \\ ? \\ ? \end{array}} & \bbox[#FF8888]{\begin{array}{c} ? \\ ? \\ ? \\ ? \\ ? \end{array}} \end{array}\!\right] \]
- Next we recall that the ones ($1$s) in the pivot columns are leading entries in their rows. Thus we place all zeros before those ones. Also we place all zeros at the bottom of the matrix. \[ \require{bbox} \left[\! \begin{array}{rrrrrrr} \begin{array}{c} \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \end{array} & \bbox[yellow]{\begin{array}{c} 1 \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \end{array}} & \begin{array}{c} \bbox[lightblue]{?} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \end{array} & \bbox[yellow]{\begin{array}{c} 0 \\ 1 \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \end{array}} & \bbox[yellow]{\begin{array}{r} 0 \\ 0 \\ 1 \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \end{array}} & \begin{array}{c} \bbox[lightblue]{?} \\ \bbox[lightblue]{?} \\ \bbox[lightblue]{?} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \end{array} & \begin{array}{c} \bbox[lightblue]{?} \\ \bbox[lightblue]{?} \\ \bbox[lightblue]{?} \\ \bbox[#88FF88]{0} \\ \bbox[#88FF88]{0} \end{array} \end{array}\!\right] \]
- That is it! The above matrix is in Reduced Row Echelon Form where $\require{bbox}\bbox[lightblue]{?}$ stands for any real number.
- You can use the above method to test whether a matrix is in RREF: I. Recognize the pivot columns. II. Test whether the pivot columns are the consecutive columns of an identity matrix. III. Test whether $1$s from the pivot columns are the leading entries in their rows. IV. All rows that do not have $1$s from the pivot columns must consist of zeros only.
- Suggested problems from Section 4.5: 12, 13, 15, 18, 19, 20
- Suggested problems for Section 4.6: 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 18, 27, 28
-
I rewrote
- Your task until I post more is to understand "An Ode to Reduced Row Echelon Form," its beauty and magic.
- Since "An Ode to Reduced Row Echelon Form" is freshly written, there are probably many errors in it. Please report all the errors that you find. I am grateful for all suggestions for improvements. I already implemented several that you communicated in class yesterday.
-
Since I often end up rewriting my writing, I invented my own saying celebrating thinking, writing, and teaching
I came, I thought, I wrote, I taught, I thought more, I rewrote.
The above words are inspired by the Latin phrase "veni, vidi, vici" which is in the list of Latin phrases very close to another Latin phrase celebrating writing:
Verba volant, scripta manent.
(Spoken words fly away, written words stay.)
-
I wrote my summary of all that we have been exploring around Reduced Row Echeon Form of a matrix. I called it
- Today we did Section 2.8 Subspaces of $\mathbb{R}^n.$ Suggested problems for Section 2.8: 5, 6, 8, 9, 10, 11-20, 24, 25, 26, 30, 31-36.
- For tomorrow, Section 2.9 Dimension and Rank. Suggested problems for Section 2.9: 1, 2, 3, 5, 6, 7, 9, 11, 12, 14, 15, 19, 20, 21, 23, 24.
- Today we did preliminaries for Section 2.3 Characterizations of invertible matrices. It is a good idea to review Exercises 23, 24, 25 from Section 2.1 before reading this section. Suggested problems for Section 2.3: 1, 3, 5, 8, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 26, 27, 33.
- Today in class we presented a different approach to the row reduction posted on Saturday. We presented that row reduction in terms of multiplication by elementary matrices.
- Recall the row reduction that we did on Saturday to find the inverse of $\displaystyle \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array}\right]$: \begin{align*} \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ -1 & 1 & 1 & 0 & 0 & 1 \end{array}\right] & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 2 & 1 & 0 & 1 & 0 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -2 & 1 & -2 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 1 & -2 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -2 & 1 & -2 \end{array}\right] \\ \end{align*}
-
Now recall that each step in row reduction can be achieved by multiplication by an elementary matrix.
Step the row operation the elementary matrix the inverse of elementary matrix 1st The third row is replaced by the the sum of the first row and the third row $E_1 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$ $E_1^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$ 2nd The third row and the second row are interchanged $E_2 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$ $E_2^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$ 3rd The third row is replaced by the the sum of the third row and the second row multiplied by $(-2)$ $E_3 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array}\right]$ $E_3^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right]$ 4th The first row is replaced by the the sum of the first row and the third row $E_4 = \left[\!\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ $E_4^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ - Next we use the elementary matrices $E_1$, $E_2$, $E_3$ and $E_4$ to reconstruct the row reduction above: \begin{align*} E_1 A & = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array}\right] = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & 0 \end{array}\right], \\ E_2 (E_1 A) & = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & 0 \end{array}\right] = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0\\ 0 & 2 & 1 \end{array}\right], \\ E_3 (E_2 E_1 A) & =\left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0\\ 0 & 2 & 1 \end{array}\right] = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right], \\ E_4 (E_3 E_2 E_1 A) & = \left[\!\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right] = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]. \end{align*}
- Because of the associativity of the matrix multiplication we have \[ (E_4 E_3 E_2 E_1) A = I_3. \]
- The reason that I wrote all these detailed steps is to show that we can write $A$ as a product of elementary matrices. Recall that \[ (E_4 E_3 E_2 E_1) A = I_3. \] Multiplying the last equality consecutively by $E_4^{-1}$, $E_3^{-1}$, $E_2^{-1}$, $E_1^{-1}$ we get \[ A = E_1^{-1} E_2^{-1} E_3^{-1} E_4^{-1}. \] Let us verify this claim: \begin{align*} E_4^{-1}& = \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right], \\ E_3^{-1} E_4^{-1} & = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right], \\ E_2^{-1} \bigl(E_3^{-1} E_4^{-1}\bigr) & =\left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right] = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & 0 \end{array}\right], \\ E_1^{-1} \bigl(E_2^{-1} E_3^{-1} E_4^{-1}\bigr) & = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & 0 \end{array}\right] = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array}\right]. \end{align*} Thus, we confirmed that \[ A = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array}\right] = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]
- The importance of the preceding item is that we wrote the matrix $A$ as a product of elementary matrices. Fore each elementary matrix we know that it is invertible. For a product of invertible matrices we know that it is invertible. In this way we have proved that $A$ is invertible just based on the fact that $A$ can be row reduced to the identity matrix.
- Since \[ A = E_1^{-1} E_2^{-1} E_3^{-1} E_4^{-1} \] and each of the matrices $E_1^{-1},$ $E_2^{-1},$ $E_3^{-1},$ $E_4^{-1}$ is (clearly) invertible, we conclude that \[ A^{-1} = E_4 E_3 E_2 E_1. \] Now calculate the product $E_4 E_3 E_2 E_1$: \begin{align*} E_1& = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right], \\ E_2 E_1 & = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right], \\ E_3 (E_2 E_1) & =\left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 1 & 0 & 1\\ -2 & 1 & -2 \end{array}\right], \\ E_4 (E_3 E_2 E_1) & = \left[\!\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & 0 \\ 1 & 0 & 1\\ -2 & 1 & -2 \end{array}\right] = \left[\begin{array}{rrr} -1 & 1 & -2 \\ 1 & 0 & 1\\ -2 & 1 & -2 \end{array}\right]. \end{align*} Thus we confirmed that \[ A^{-1}= \left[\begin{array}{rrr} -1 & 1 & -2 \\ 1 & 0 & 1\\ -2 & 1 & -2 \end{array}\right] = E_4 E_3 E_2 E_1. \]
-
After reading this post you should be able to solve an exercise stated like this:
Consider the matrix $M = \left[\begin{array}{rrr} 3 & 3 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 0 \end{array}\right]$.
- Determine whether it is possible to write the matrix $M$ as a product of elementary matrices.
- If you claim that it is possible to write $M$ as a product of elementary matrices, then find elementary matrices whose product is $M$. If you claim that it is not possible to write $M$ as a product of elementary matrices, justify your answer.
- Yesterday we continued with Section 2.2: The inverse of a matrix. We proved Theorem 4, Theorem 5, and the first part of Theorem 7.
- It is important for you to perfect calculation of the inverse of an invertible matrix. For a $2\!\times\!2$ matrix one should use Theorem 4. For a $3\!\times\!3$ matrix one uses the method presented in Example 7.
-
Consider the following exercise:
- Determine whether the matrix $A = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array}\right]$ is invertible.
- If $A$ is invertible find its inverse $A^{-1}$.
-
To answer the fist question in the exercise we use the implication
If RREF of $A$ is $I_3$, then $A$ is invertible.
This implication is proved in Theorem 7 in Section 2.2 . This proof is important! - So, we just row reduce $A$ to RREF and that will answer the first question in the exercise: \begin{align*} \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array}\right] & \sim \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & 0 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \end{align*}
- The row reduction above shows that $A$ is row equivalent to $I_3$. Therefore, by Theorem 7 in Section 2.2 , $A$ is invertible.
- To calculate the inverse $A^{-1}$ we row reduce the $3\times 6$ matrix $[A | I_3]$: \begin{align*} \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ -1 & 1 & 1 & 0 & 0 & 1 \end{array}\right] & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 2 & 1 & 0 & 1 & 0 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -2 & 1 & -2 \end{array}\right] \\ & \sim \left[\!\begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 1 & -2 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -2 & 1 & -2 \end{array}\right] \\ \end{align*}
- The matrix whose inverse we found above is a very special matrix. It is a matrix whose entries are integers and all the entries of its inverse are integers. In exercises it is often convenient to have such matrices: matrices with integer entries whose inverse is also a matrix with integer entries. Such matrices are called unimodular matrices. I was surprised that there are a lot of such matrices. Here is a pdf file with all unimodular matrices with entries $-1,0,1,2,3$ whose inverses have entries among the ten digits and their opposites. I have omitted the matrices with 3 or more zeros. This pdf file is huge, 3119 pages with over 80000 matrices. If you ever need a unimodular matrix I hope you find it here. You can use these matrices to practice row reduction since for every matrix listed in this file I list its inverse.
- Today we started Section 2.2: The inverse of a matrix; suggested problems are: 1, 4, 5, 7, 12, 13, 21, 22, 23, 24, 28, 32, 33, 34, 38. In all bold face problems the textbook asks you to "explain why" something is true. In the text of the book and in class we offer proofs. Please try to organize your solutions as proofs, not explanations. State clearly what is known from theorems in the book and use logical reasoning to deduce what the problem claims.
-
My approach to the inverse of a $2\!\times\!2$ matrix is a little different from the approach in the textbook. I look at the process of seeking: Given a $2\!\times\!2$ matrix $\displaystyle \left[\!\begin{array}{cc}
a & b \\ c & d \end{array}\!\right]$ we seek another $2\!\times\!2$ matrix $\displaystyle \left[\!\begin{array}{cc}
x & y \\ z & w \end{array}\!\right]$ such that
\[
\left[\!\begin{array}{cc}
a & b \\ c & d \end{array}\!\right]\left[\!\begin{array}{cc}
x & y \\ z & w \end{array}\!\right] = \left[\!\begin{array}{cc}
ax+bz & ay+bw \\ cx+dz & cy+dw \end{array}\!\right] = \left[\!\begin{array}{cc}
1 & 0 \\ 0 & 1 \end{array}\!\right].
\]
At the first step we focus on the zeros and seek $y$ and $w$ such that $ay+bw = 0$, and $x$ and $z$ such that $cx+dz = 0.$
One of the simplest solution of $ay+bw = 0$ is $a(-b)+b\,a = 0$ and one of the simplest solution of $cx+dz = 0$ is $c\,d+d(-b) = 0.$ (This might take some guessing since you have another choice of $c(-d)+d\,b = 0.$) Now try $\displaystyle \left[\!\begin{array}{cc}
d & -b \\ -c & a \end{array}\!\right]$:
\[
\left[\!\begin{array}{cc}
a & b \\ c & d \end{array}\!\right]\left[\!\begin{array}{cc}
d & -b \\ -c & a \end{array}\!\right] = \left[\!\begin{array}{cc}
ad-bc & 0 \\ 0 & c(-b)+da \end{array}\!\right] = (ad-bc) \left[\!\begin{array}{cc}
1 & 0 \\ 0 & 1 \end{array}\!\right].
\]
From the last equality we see that if $ad-bc \neq 0$, then the matrix $\displaystyle \left[\!\begin{array}{cc}
a & b \\ c & d \end{array}\!\right]$ is invertible and its inverse is
\[
\displaystyle \left[\!\begin{array}{cc}
a & b \\ c & d \end{array}\!\right]^{-1} = \frac{1}{ad-bc}\left[\!\begin{array}{cc}
d & -b \\ -c & a \end{array}\!\right].
\]
My point here is that you do not have to memorize the inverse, you can construct it on your own just playing with the equations $ay+bw = 0$ and $cx+dz = 0$ and their simple solution.
The number $ad-bc$ is called the determinant of the matrix $\displaystyle \left[\!\begin{array}{cc} a & b \\ c & d \end{array}\!\right]$. We write \[ \det \left[\!\begin{array}{cc} a & b \\ c & d \end{array}\!\right] = \left|\!\begin{array}{cc} a & b \\ c & d \end{array}\!\right| = ad-bc. \]
- Yesterday and today we did Section 2.1, I would call it Algebra with Matrices. Suggested problems are: 1, 4, 5, 7, 9, 10, 11, 12, 17, 20, 22, 23, 24, 25, 27, 28, 34. The jewel of this section is Matrix Multiplication. It is such a beautiful and rich concept with three distinct ways of looking at it. I talked about it in class today and last week, so please check the class notes on Canvas.
- Please pay special attention to problems 22, 23, 24, 25 from Section 2.1. In each of these problems you are asked to prove an implication.
-
For each of these problems, and in general whenever you try to prove something, you should rewrite the statement of the problem as a clear implication. That is, as a statement of the form
- Recall that the converse of the implication $p \Rightarrow q$ is the implication $q \Rightarrow p$. Also recall that the converse can be wrong when the original proposition is true.
- For Problem 22 in Section 2.1 write down the converse and answer whether it is true or false. Justify your answer.
- For Problem 24 in Section 2.1 write down the converse and answer whether it is true or false. Justify your answer. This problem is closely related to Problem 26 in Section 2.1.
- Today I presented how linear algebra is useful in computer graphics. A part of that is understanding the linear algebra aspects of the Color Cube.
- Today we started Section 2.1: Matrix operations. For now, the objective is to learn matrix multiplication and apply it to verify row reduction. Suggested problems are: 1, 5, 6, 7, 8, 9, 10, 11, 12.
-
Below is my attempt to use colors to explain how multiplication of matrices works:
-
In the three pages below I summarize the content from the last homework.
- We did Section 1.8 and Section 1.9 on Friday and today.
- Suggested problems for Section 1.8: 1-4, 12, 13-17, 19, 20, 25, 27, 28
- Suggested problems for Section 1.9: 1, 3, 4, 5, 7, 8, 11, 12, 18, 19, 23
- In the textbook the authors list several reflections transformations across some popular lines like: $x_2 = 0$ (this is $x_1$-axis), $x_1 = 0$ (this is $x_2$-axis), $x_2 = x_1$ (this is the diagonal of the first and the third quadrant), and $x_2 = -x_1$ (this is the diagonal of the second and the fourth quadrant). But why not establish the standard matrix of the reflection across the line which makes the angle $\theta$ with the positive $x_1$-axis? It is not more difficult than the standard matrix of the rotation by the angle $\theta$ in counterclockwise direction, which is \[ \left[\! \begin{array}{cc} \cos(\theta) & -\sin(\theta) \\[5pt] \sin(\theta) & \cos(\theta) \end{array}\!\right] \]
-
The picture below will help us determine the standard matrix of the reflection across the blue line which makes angle $\theta$ with the positive $x_1$-axis. Denote this reflection by
\[
F:\mathbb{R}^2 \to \mathbb{R}^2.
\]
To determine the standard matrix of this reflection we need to calculate the coordinates of the vectors
\[
F \mathbf{e}_1 = F \left[\! \begin{array}{c}
1 \\[3pt]
0
\end{array}\!\right] \quad \text{and} \quad F \mathbf{e}_2 = F \left[\! \begin{array}{c}
0 \\[3pt]
1
\end{array}\!\right]
\]
In the picture below the vector $F \mathbf{e}_1$ is dark orange and the vector $F \mathbf{e}_2$ dark purple.
- By the definition of the reflection $F$, the angle formed by the light orange coordinate vector $\mathbf{e}_1$ and its reflection the dark orange vector $F \mathbf{e}_1$ is $2\theta.$ Therefore, \[ F \mathbf{e}_1 = F \left[\! \begin{array}{c} 1 \\[3pt] 0 \end{array}\!\right] = \left[\! \begin{array}{c} \cos(2\theta) \\[3pt] \sin(2\theta) \end{array}\!\right]. \]
- In this item we calculate the coordinates of the dark purple vector $F \mathbf{e}_2$ which is the reflection of the light purple coordinate vector $\mathbf{e}_2.$ First, we observe that the light gray right triangles in the picture below are congruent. This claim follows from the fact that the hypothenuse of each of these triangles is of length $1$ and the triangles have the same angle $2\theta$ at the origin. That the angle at the origin of the top light gray triangle is $2\theta$ follows from the definition of the reflection. To calculate the angle at the origin for the lower light gray triangle we calculate the angle between the light purple coordinate vector $\mathbf{e}_2$ and its reflection, the dark purple vector $F \mathbf{e}_2;$ since the angle between the light purple coordinate vector $\mathbf{e}_2$ and the blue line is $\displaystyle \frac{\pi}{2} - \theta$, the angle between is $\mathbf{e}_2$ and $F \mathbf{e}_2$ is \[ 2 \left(\frac{\pi}{2} - \theta\right) = \pi - 2 \theta. \] Since the sum of this angle and the angle at the origin for the lower light gray triangle equals $\pi,$ we have proven that the angle at the origin for the lower light gray triangle equals $2\theta.$ In the preceding item we proved that the horizontal side of the top light gray right triangle is $\cos(2\theta)$ and its vertical side is $\sin(2\theta)$. As a consequence, the horizontal side of the lower light gray right triangle is $\sin(2\theta)$ and its vertical side is $\cos(2\theta)$. Adjusting for the positioning in the coordinate system we get \[ F \mathbf{e}_2 = F \left[\! \begin{array}{c} 0 \\[3pt] 1 \end{array}\!\right] = \left[\! \begin{array}{c} \sin(2\theta) \\[3pt] -\cos(2\theta) \end{array}\!\right]. \] Finally we have that the standard matrix of the reflection across the blue line which makes angle $\theta$ with the positive $x_1$-axis is \[ \left[\! \begin{array}{cc} \cos(2\theta) & \sin(2\theta) \\[5pt] \sin(2\theta) & -\cos(2\theta) \end{array}\!\right]. \] Enjoy Reflections!
Enjoy Reflections!
-
Problems 41, 42, 43, 44 in Section 1.7: Linear independence are very interesting and important. The matrices in these problems are not easy to row reduce by hand, so the textbook recommends that we use a calculator. Here I am offering a simpler matrix and more detailed guidance for the calculations. This is my version of these four problems.
Problem. Consider the following $4\times 6$ matrix \[ A = \left[ \begin{array}{cccccc} 1 & 1 & 4 & 1 & 0 & 6 \\ 2 & 1 & 3 & 0 & 1 & 4 \\ 3 & 1 & 2 & 1 & 0 & 8 \\ 4 & 1 & 1 & 0 & 1 & 6 \\ \end{array} \right]. \] Denote the columns of $A$ by $\mathbf{a}_1,$ $\mathbf{a}_2,$ $\mathbf{a}_3,$ $\mathbf{a}_4,$ $\mathbf{a}_5,$ and $\mathbf{a}_6.$ That is, \[ \mathbf{a}_1 = \left[ \begin{array}{cccccc} 1 \\ 2 \\ 3 \\ 4 \\ \end{array} \right], \quad \mathbf{a}_2 = \left[ \begin{array}{c} 1\\ 1 \\ 1 \\ 1 \\ \end{array} \right], \quad \mathbf{a}_3 = \left[ \begin{array}{c} 4 \\ 3 \\ 2 \\ 1 \\ \end{array} \right], \quad \mathbf{a}_4 = \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \\ \end{array} \right], \quad \mathbf{a}_5 = \left[ \begin{array}{c} 0\\ 1\\ 0\\ 1 \\ \end{array} \right], \quad \mathbf{a}_6 = \left[ \begin{array}{c} 6 \\ 4 \\ 8 \\ 6 \\ \end{array} \right]. \] Answer the following questions:
- Find the Reduced Row Echelon Form of $A$. (Please make sure that it is done correctly.)
- Is the vector $\mathbf{a}_1$ linearly independent? Explain using the definition.
- Are the vectors $\mathbf{a}_1,$ $\mathbf{a}_2$ linearly independent? Explain using the definition.
- Are the vectors $\mathbf{a}_1,$ $\mathbf{a}_2,$ $\mathbf{a}_3$ linearly independent? Explain using the definition. If the answer is No, then express $\mathbf{a}_3$ as a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2.$ That is, find the real numbers $\alpha_1$ and $\alpha_2$ such that \[ \mathbf{a}_3 = \alpha_1 \mathbf{a}_1+ \alpha_2 \mathbf{a}_2. \] Make a record of the numbers $\alpha_1$ and $\alpha_2.$
- Are the vectors $\mathbf{a}_1,$ $\mathbf{a}_2,$ $\mathbf{a}_4$ linearly independent? Explain using the definition.
- Are the vectors $\mathbf{a}_1,$ $\mathbf{a}_2,$ $\mathbf{a}_4,$ $\mathbf{a}_5$ linearly independent? Explain using the definition. If the answer is No, then express $\mathbf{a}_5$ as a linear combination of $\mathbf{a}_1,$ $\mathbf{a}_2$ and $\mathbf{a}_4.$ That is, find the real numbers $\beta_1,$ $\beta_2$ and $\beta_4$ such that \[ \mathbf{a}_5 = \beta_1 \mathbf{a}_1+ \beta_2 \mathbf{a}_2 + \beta_4 \mathbf{a}_4. \] Make a record of the numbers $\beta_1,$ $\beta_2$ and $\beta_4.$
- Are the vectors $\mathbf{a}_1,$ $\mathbf{a}_2,$ $\mathbf{a}_4,$ $\mathbf{a}_6$ linearly independent? Explain using the definition. If the answer is No, then express $\mathbf{a}_6$ as a linear combination of $\mathbf{a}_1,$ $\mathbf{a}_2$ and $\mathbf{a}_4.$ That is, find the real numbers $\gamma_1,$ $\gamma_2$ and $\gamma_4$ such that \[ \mathbf{a}_6 = \gamma_1 \mathbf{a}_1+ \gamma_2 \mathbf{a}_2 + \gamma_4 \mathbf{a}_4. \] Make a record of the numbers $\gamma_1,$ $\gamma_2$ and $\gamma_4.$
- The moral of this problem is that all the information you found about the linear independence and dependence of the columns of $A$ is contained in the Reduced Row Echelon Form of $A$. I would like you to carefully review your findings and discover how are all the facts about the columns of $A$ that you found in this problem encoded in the Reduced Row Echelon Form of $A$.
- We started Section 1.7: Linear independence today. Suggested problems: 2, 4, 5, 8, 10, 11, 17, 21, 23, 24, 25, 26, 27, 28, 29, 32, 33, 34, 35, 36, 37, 38, 39, 40
- We skipped Section 1.6: Applications of Linear Systems.
- Today we discussed Section 1.5 which talks about writing solution sets of linear systems in parametric vector form. We explained the relationship between the solution set of the homogeneous equation $A \mathbf x = \mathbf 0$ and the solution set of a consistent nonhomogeneous equation $A \mathbf x = \mathbf b$; see Theorem 6 for details. Please recognize how this theorem is reflected when the solution of $A \mathbf x = \mathbf 0$ is written in parametric vector form. Suggested problems for Section 1.5: 1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 16, 19, 21, 23, 24, 26, 29, 32, 35, 37, 38, 39, 40. While working on these problems, try to recognize a span of one, or two, or three vectors when you write a formula for the solution in parametric form.
- Today we discussed Section 1.4: The Matrix Equation $A\mathbf{x} = \mathbf{b}$. Suggested problems for Section 1.4: 1, 5, 13, 14, 15, 16,17-20, 21, 22, 23, 24, 25, 26, 35, 36 and problems added in this post 32 and 34.
- Today we discussed Section 1.3: Vector Equations. Suggested problems for Section 1.3: 1, 5, 9, 15, 17, 18, 19, 21-25, 32.
- One typical question that is asked in this section is: Is the vector $\left[\begin{array}{c} 3 \\ 1 \\ 2 \end{array} \right]$ in the span of the vectors $\left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right]$ and $\left[\begin{array}{c} 2 \\ 3 \\ 1 \end{array} \right]$? Or, the same question in more mathematical notation: Is the following statement true \[ \left[\begin{array}{c} 3 \\ 1 \\ 2 \end{array} \right] \in \operatorname{Span}\left\{ \left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right], \left[\begin{array}{c} 2 \\ 3 \\ 1 \end{array} \right] \right\}? \] What is really interesting about linear algebra this question is just a linear system in disguise. In fact, we are asking if the following vector equation has a solution \[ x_1 \left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right] +x_2 \left[\begin{array}{c} 2 \\ 3 \\ 1 \end{array} \right] = \left[\begin{array}{c} 3 \\ 1 \\ 2 \end{array} \right]. \] In turn, this vector equation is in fact a system of linear equations, like we did on the first day of classes: \begin{alignat*}{7} &x_1 & & + 2 &x_2 & &= 3\\ 2 &x_1 & & + 3 &x_2 & & = 1 \\ 3 &x_1 & & + &x_2 & & = 2 \end{alignat*} This system is solved by row reducing the augmented matrix: \begin{align*} \left[\!\begin{array}{cc|c} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\! \right] & \sim \left[\!\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & -1 & -5 \\ 0 & -5 & -7 \end{array}\! \right] \\ & \sim \left[\!\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 1 & 5 \\ 0 & -5 & -7 \end{array}\! \right] \\ & \sim \left[\!\begin{array}{cc|c} 1 & 0 & -7 \\ 0 & 1 & 5 \\ 0 & 0 & 18 \end{array}\! \right] \\ & \sim \left[\!\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\! \right] \quad \small{\text{I always like to row reduce a matrix to its RREF.}} \end{align*} Since the augmented column of the RREF of the augmented matrix is a pivot column the given system is inconsistent; it does not have a solution. Thus, the answer to the given question is no! The given vector is not in the span of the given two vectors.
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The question which was answered in the previous item is typical of this section. I am writing this to emphasize that a very similar question is not asked in this section. The question is:
- Today we talked about Section 1.2. Suggested problems for Section 1.2: 3, 5, 6, 7, 8, 12, 17-31.
- In addition to the above problems in Section 1.2 do Problem 33.
- Also, in addition to the problems suggested for Section 1.1 do Problem 34.
- A useful link is Wikipedia's Row echelon form.
- A matrix whose all entries are zero is called a zero matrix. A row of a matrix is said to be a zero row if all the entries in that row are zero. The leftmost nonzero entry of a nonzero row is called the leading entry of a row. The zeros preceding the leading entry are called the leading zeros of a row. All entries of a zero row are leading zeros.
- This is a restatement of Wikipidia's definition of reduced row echelon form:
Each zero matrix is in reduced row echelon form. A nonzero matrix is in reduced row echelon form if it satisfies the following three conditions:
- Each nonzero row has strictly more leading zeros then the row preceding it.
- The leading entry of each nonzero row is equal to $1$.
- The leading entry of each nonzero row is the only nonzero entry in its column.
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To illustrate the definition of the reduced row echelon form made a picture of a $12\times 25$ matrix (that is a matrix with $12$ rows and $25$ columns which is in reduced row echelon form. To emphasize the role of the leading zeros, I colored them blue. To emphasize the role of leading $1$s, I colored them green. I also colored in green the each pivot column. The entries of the matrix that are not specifically written as $0$s or $1$s can be any real numbers. Please notice how blue leading zeros form steps which climb from right to left. Also notice that each stair hosts a pivot position, that is the leading $1$ in a nonzero row.
- A very good exercise in understanding RREF (reduced row echelon form) is to write down all $2\!\times\!3$ matrices which are in RREF. There are seven of them. Below I list the matrix with no pivot columns, then all matrices with one pivot column, then the matrices with two pivot columns. Those are all possibilities since each pivot column has pivot position occupied by the leading entry which needs its own row and we are studying matrices with only two rows. In the matrices below stars ($*$) stand for arbitrary real numbers. \begin{align*} & \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ & \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix},\quad \begin{bmatrix} 0 & 1 & * \\ 0 & 0 & 0 \end{bmatrix},\quad \begin{bmatrix} 1 & * & * \\ 0 & 0 & 0 \end{bmatrix},\\ & \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & * & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 & * \\ 0 & 1 & * \end{bmatrix}, \quad \end{align*}
- Here are all possible $3\!\times\!4$ matrices in RREF (reduced row echelon form). There are fifteen of them. \begin{align*} & \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \\ & \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \begin{bmatrix} 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix},\ \begin{bmatrix} 0 & 1 & * & * \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \begin{bmatrix} 1 & * & * & * \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix},\\ & \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \begin{bmatrix} 0 & 1 & * & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \begin{bmatrix} 1 & * & * & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \begin{bmatrix} 0 & 1 & 0 & * \\ 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \begin{bmatrix} 1 & * & 0 & * \\ 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \begin{bmatrix} 1 & 0 & * & * \\ 0 & 1 & * & * \\ 0 & 0 & 0 & 0 \end{bmatrix}, \\ & \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \ \begin{bmatrix} 1 & * & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},\ \begin{bmatrix} 1 & 0 & * & 0 \\ 0 & 1 & * & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \ \begin{bmatrix} 1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & * \end{bmatrix}, \end{align*}
- Here are all possible $4\!\times\!3$ matrices in RREF (reduced row echelon form). There are eight of them. \begin{align*} & \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ & \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 0 & 1 & * \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix},\quad \begin{bmatrix} 1 & * & * \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ & \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 1 & * & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 & * \\ 0 & 1 & * \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}. \end{align*}
- Notice that in all the matrices listed in the preceding item the bottom row is the zero row. By removing that zero row we would get all possible $3\!\times\!3$ matrices in RREF (reduced row echelon form). Thus, there is eight different types of $3\!\times\!3$ matrices in RREF (reduced row echelon form). This is true in general: If $m\geq n$, than the number of different types of $m\!\times\!n$ matrices in RREF is the same as the number of different types of $n\!\times\!n$ matrices in RREF.
- Another good exercise is to think of the above matrices as augmented matrices of systems of equations and state for each corresponding system whether it has: No Solutions (NS), Unique Solution (US), Infinitely Many Solutions (MS).
- Another interesting exercise, but for Math 309, is to calculate how many $m\!\times\!n$ matrices are in RREF. Just for the record: Let $k$ be a nonnegative integer and let $m, n$ be positive integers such that $k \leq m \leq n.$ It turns out that there are exactly \[ \displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!} \] matrices of size $m\!\times\!n$ with $k$ pivots (leading $1$s). The symbol $\displaystyle \binom{n}{k}$ is the standard notation for the binomial coefficient $n$-choose-$k$. Since an arbitrary $m\!\times\!n$ matrix can have $k \in \{0,1,\ldots,m\}$ pivots, it follows that the number of possible RREF types for an $m\!\times\!n$ matrix is \[ \sum_{k=0}^m \binom{n}{k}. \] For example, with $m=3$ and $n=4$, as seen above, there are \[ \binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} = 1 + 4 + 6 + 4 = 15. \] possible RREF types of matrices. It is interesting to observe that for an $n\!\times\!n$ matrix there are \[ \sum_{k=0}^n \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n-1} + \binom{n}{n} = 2^n \] possible RREF types. If you take Math 309 you will understand the background of these formulas.
- In conclusion, for an $m\!\times\!n$ matrix with $m\lt n$ there are \[ \sum_{k=0}^m \binom{n}{k} \] possible RREF types and if $m \geq n$ there are \[ 2^n \] possible RREF types.
- I recommend that you look into learning LaTeX which is a superior typesetting system for writing math papers. I created Getting Started with LaTeX page to help you with this.
- Here is a simple LaTeX sample document in which I prove an interesting inequality.
- Here is a a LaTeX assignment template which you can use to write your assignments.
- If you need help in starting with LaTeX feel free to ask me for help. I consider it as important as learning one nice piece of mathematics. And I want to help you with that, not only because that is my job, but more because I deeply believe that both math - through learning rigorous thinking - and professional writing skills - supported by rigorous thinking - will help you in your life more than anything else.
- The information sheet
- We will start with Section 1.1 Systems of Linear Equations. Suggested problems are 3, 6, 7, 5, 9, 11, 12, 16, 17, 18, 21, 22, 23, 24, 25, 27, 31, 33.
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Just reading Section 1.1 you will experience that linear algebra is a subject full of new terminology:
- linear equation,
- coefficients of a linear equation,
- system of linear equations or linear system or just system,
- a solution of a linear system,
- the solution set of a linear system,
- consistent linear system,
- inconsistent linear system,
- equivalent systems,
- coefficient matrix of a system,
- augmented matrix of a system,
- elementary row operations,
- row-equivalent matrices.
- Learning the terminology of a field is crucial to getting to understand that field. This is true in general and particularly true in Linear Algebra. Therefore I started a webpage with Glossary of Linear Algebra Terms with my comments.
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What is the oldest linear algebra problem?
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Clay tablet VAT 8389 from the Old Babylonian period, from 2000 to 1600 BC, contains what is believed to be the earliest word problem that can be interpreted as a system of linear equations:
A translation of this word problem into a system of linear equations is as follows: \begin{alignat*}{4} &x_1 & &\ + &x_2 & = 1800 \\ \tfrac{2}{3} &x_1 & &- \tfrac{1}{2} &x_2 & = \phantom{1}500. \end{alignat*} -
Problem 40 of the Rhind papyrus which is dated to 1550 BC is:
Denote by $x_1$ the smallest number and by $x_2$ the common difference. After simplification the above problem translates into the following system of linear equations: \begin{alignat*}{5} 5 &x_1 & & + 10 &x_2 & = 100 \\ \tfrac{11}{7} &x_1 & & - \phantom{1}\tfrac{2}{7} &x_2 & = \phantom{10}0. \end{alignat*} -
Most importantly for us, the oldest known treatment of systems of linear equations from antiquity which resembles the methods that we will use in this class is in Chapter 8 of the Chinese textbook Nine Chapters of the Mathematical Art which is at least 1800 years old.
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Clay tablet VAT 8389 from the Old Babylonian period, from 2000 to 1600 BC, contains what is believed to be the earliest word problem that can be interpreted as a system of linear equations: