- I updated the list of topics that we covered during the quarter.
- Assigned exercises for Section 5.3 2, 3, 5, 8, 9, 12, 13, 16, 18, 20, 23, 24.
- Assigned exercises for Section 5.2 are 1-8 (find corresponding eigenvectors as well), 9, 13, 15, 18, 19, 20, 21, 24, 25, 27.
- In class I did a different proof of very important Theorem 2 in Section 5.1. In class I did the case of two vectors. Here I post the case of three vectors. This will give you the idea how to do any number of vectors. In the book they do a proof by contradiction; I prefer a direct proof. I hope that studying both proofs will make it easier for you to internalize this proof.
- 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.
- Here are animations of different matrices in action. In each scene the navy blue vector is the image of the sea green vector under multiplication by a matrix A. For easier visualization of the action the heads of the 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.
|
|
|
|
|
|
|
|
- I updated the list of topics that we covered so far.
- Suggested problems for Section 4.7: 4-10, 13, 14
- Suggested problems for Section 4.6: 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 18, 23, 24, 27, 28, 29, 30 (there is a typo in this problem; the vector $\mathbf b$ should be in ${\mathbb R}^m$).
- Suggested problems for Section 4.5: 3, 6, 7, 8, 9, 10, 12, 13, 15, 18, 19, 20, 21, 23
- Suggested problems for Section 4.4: 3, 4, 7, 8, 9, 10, 11, 12, 13, 14, 27, 28
- A natural extension of Section 4.4 is Section 4.7. Based on what we did in Section 4.4 you should be able to do exercises 7, 8, 9, 10 in Section 4.7.
- Suggested problems for Section 4.3: 3, 4, 5, 9, 10, 11, 13, 14, 15, 21, 22, 25, 33, 34
- Suggested problems for Section 4.2: 1, 3, 7, 9, 12, 15, 17, 18, 19, 23, 24, 25, 26, 28, 31, 32, 33
- I updated the list of topics that we covered so far.
- After the exam you can do problems for Section 4.1: 1, 3, 7, 8, 9, 11 13, 15, 17, 21, 23, 24
- Suggested problems for Section 3.2: 5, 7, 9, 11, 16-20 (even), 21, 25, 31, 33, 34, 35, 40c
- Suggested problems for Section 3.1: 1, 3, 5, 25-30, 12, 13, 37, 40, 41
- 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.
- 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.
-
After internalizing yesterday's post you should be able to solve an exercise stated like this:
- Determine whether it is possible to write the matrix $M = \left[\begin{array}{rrr} 3 & 3 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 0 \end{array}\right]$ 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, then justify your claim.
- Suggested problems for Section 2.4: 3, 4, 5, 6, 10.
-
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 equivalence (a) $\Leftrightarrow$ (b) in the Invertible Matrix Theorem. The equivalence (a) $\Leftrightarrow$ (b) in the Invertible Matrix Theorem reads:
$A$ is invertible if and only if RREF of $A$ is $I_3$.
- 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, $A$ is invertible.
-
Now recall that each step in the row reduction can be achieved by multiplication by an elementary matrix.
Step the row operation the elementary matrix the inverse of ele mat 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 product we have \[ (E_4 E_3 E_2 E_1) A = I_3. \] Since we already know that $A$ is invertible, the last equality shows 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 calculated \[ A^{-1}= \left[\begin{array}{rrr} -1 & 1 & -2 \\ 1 & 0 & 1\\ -2 & 1 & -2 \end{array}\right]. \]
- A more direct way of calculating $A^{-1}$ is to 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*} Notice that in the above row reduction, the matrix $I_3$ is being transformed to $E_1$ at the 1st step, to $E_2 E_1$ at the 2nd step, to $E_3 (E_2 E_1)$ at the 3rd step and finally to $E_4 (E_3 E_2 E_1) = A^{-1}$ at the 4th step.
- The reason that I wrote all these detailed steps is to show that we obtained $A^{-1}$ as a product of elementary matrices. It is interesting to point out that at the same time we obtained $A$ 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 \[ A = E_1^{-1} E_2^{-1} E_3^{-1} E_4^{-1}. \] Thus, \[ A = \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array}\right] = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] It is a good exercise in matrix arithmetic to confirm that the last equality is correct.
-
After reading this post you should be able to solve an exercise stated like this:
- Determine whether it is possible to write the matrix $M = \left[\begin{array}{rrr} 3 & 3 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 0 \end{array}\right]$ as a product of elementary matrices.
- If it is possible to write $M$ as a product of elementary matrices 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.
- I just started Section 2.3. It is a good idea to read it and try some problems. Reviewing Exercises 23, 24, 25 from Section 2.1 is also a good idea. Suggested problems for Section 2.3: 1, 3, 5, 8, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 26, 27, 33.
- Notice that I did in detail Exercises 23, 24, 25 from Section 2.1.
- Suggested problems for Section 2.2: 1, 4, 5, 13, 21, 22, 23, 28, 31, 33, 34, 38.
- Suggested problems for Section 2.1: 1, 3, 5, 7, 9, 10, 11, 17, 20, 22, 23, 24, 25, 27, 28, 34
- I updated the list of topics that we covered so far.
- Suggested problems for Section 1.8: 1-4, 12, 13-16, 17, 27, 28
- Suggested problems for Section 1.9: 1, 3, 4, 5, 7, 8, 11, 12, 18, 19, 23, 25, 26, 27, 28, 29, 30
- Here is the file I which I use matrices to manipulate a picture of my dog. The name of the file is Pictures.nb. The jpg picture that I manipulate in this file is here. To make the Mathematica file Pictures.nb work you need to save it to a directory which full path you know. Save the picture in the same directory. Then change the Mathematica working directory in Pictures.nb to the directory to which you downloaded the file and the picture. The post on April 8, 2014 has some information how to get started with Mathematica.
- Here is a list of topics that we covered so far. I will keep this list updated so that we have a list of all topics that will be covered on the final.
- I updated the list of 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.
- Suggested problems for Section 1.7: 2, 4, 5, 8, 10, 11, 17, 21, 23, 24, 25, 26, 27, 28, 29, 32, 33, 34, 35, 36, 37, 38, 39, 40
- 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.
- Suggested problems for Section 1.4: 1, 5, 13, 14, 15, 16, 17-20, 21, 22, 23, 24, 25, 26, 35, 36
- Here is the list all possible $3\times 4$ matrices in RREF (row reduced 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}, \quad \begin{bmatrix} 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix},\quad \begin{bmatrix} 0 & 1 & * & * \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \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}, \quad \begin{bmatrix} 0 & 1 & * & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 1 & * & * & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 0 & 1 & 0 & * \\ 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 1 & * & 0 & * \\ 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \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}, \quad \begin{bmatrix} 1 & * & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},\quad \begin{bmatrix} 1 & 0 & * & 0 \\ 0 & 1 & * & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & * \end{bmatrix}, \end{align*}
- To save time, today in class I used Mathematica to row reduce a matrix.
Here is the simple Mathematica file that I created. It is called 20140408.nb. Right-click on the underlined word "Here"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 5.2. You will find Mathematica 5.2 on all campus computers in
Start -> All Programs -> Math Applications -> Mathematica. Open Mathematica first, then open 20140408.nb from Mathematica. - If you have problems running files that I post here please let me know. If you spend some time learning how to use Mathematica you will enhance the understanding of math that we are studying.
- We also have Mathematica version 8. It is only available in BH 215. Since Mathematica v 5.2 is more widely available I decided to use it in this class. These two versions are not compatible. The command structure is very similar. Version 8 will usually recognize the differences and correct them.
- More information on how to use Mathematica you can find on my Mathematica v5.2 page and Mathematica v8 page.
- Suggested problems for Section 1.3: 1, 5, 9, 15, 17, 18, 19, 21-25, 32
- As I pointed out in class, a very good exercise in understanding RREF (row reduced echelon form) is to write down all $2\times 3$ matrices which are in RREF. There are seven of them. \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*}
- 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, Unique Solution, Infinitely Many Solutions.
- One could repeat the above two bullets for $3\times 3$ and $3\times 4$ matrices.
- Read Section 1.2. Suggested problems for Section 1.2: 3, 5, 6, 7, 8, 12, 17-31
- Some useful links:
- 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 entries in that row are zero. The leftmost nonzero entry of a nonzero row is called a leading entry. The zeros preceding the leading entry are called leading zeros of a row. All entries of a zero row are leading zeros.
- This is a restatement of PlanetMath's definition of row echelon form:
A nonzero matrix is said to be in row echelon form if its first row is a nonzero row and each nonzero row bellow the first has strictly more leading zeros then the previous row.
- This is a restatement of Wikipidia's definition of reduced row echelon form:
A nonzero matrix which is in row echelon form is said to be in reduced row echelon form if the leading entries off all nonzero rows are equal to 1 and this 1 is the only nonzero entry in its column.
- 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.