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and...matrixes to a power

I’m working on the practice test and there are just two things that I need to clear up before I feel confident to take the test. I’m sorry for this email being so late at night, but I can also drop in on your office hours tomorrow before the test if that works better for you.

On the practice test question 2 part c I wanted to know if there was an easier way to prove that the inverse of B had the same eigenvectors as B and the eigenvalues are the inverse of B’s eigenvalues besides the time consuming way to invert the matrix the recalculate the eigenvector and eigenvalues. 

Now that the matrix is invertible, note that x^(-1) is a power of x and we talked about how to easily compute the power of a matrix using the diagonalization? 

Or would it not matter at all since the matrix was singular and cannot be inverted. And for part d it sounds like you’re asking for the linear combination, but does the phrase “expand the vector v in terms of the normalized eigenvectors of B” change that?

Expand the vector in terms of the normalized eigenvectors means exactly asking for the coefficients needed to make the linear combination

I also think that the third question is asking for the linear combination of both basis’, but if it is then I don’t know how I would use the answer for a to solve b since you could just solve it twice. 

for part b) I'm asking you to go from the linear combination in terms of one basis into the linear combination in terms of the other basis, remember how we did this in class? 

I’m mostly concerned that I don’t recognize the terminology for most of the test and I can’t find a place that lists all the ways to say them. Which is especially concerning because this is the third time I found the linear combination while working on the practice test and none of them are worded the same way, and I just want to make sure I’m doing this right before I take the test.

**********************

5.1#12

Hello Dr. Taylor, 

I was having trouble with the Gram-Schmidt procedure in the homework and I wanted to check the practice test too. I found that the first vector will be the vector, the second would be the second (v2) – dot product of v1 and v2 divided by v1Tv1 times v1 and the third is the same process but also have v3 -  dot product of v1 and v3 divided by v1Tv1 times v1 - dot product of the original v2 and v3 divided by v2Tv2 times v2. 

Indeed, start with v1. BUT, you're supposed to give a normalized vector, so your first vector should be u1 = v1/||v1||

 v2– (v1T v2)/( v1Tv1) v1 will indeed be perpendicular to v1 (and this is the same as v2-(u1^Tv2)u1

because u1Tu1=1) It will not typically be normalized though, except by accident. You need to normalize it: u2 = (v2– (v1T v2)/( v1Tv1) v1)/||(v2– (v1T v2)/( v1Tv1) v1)||.

Now, v3 - (u1Tv3)u1 - (u2^Tv3)u2 will be perpendicular to both v1 and v2, and to both u1 and u2. Again, it's not normalized so you need to do that.

For the first problem I got the second row right and the first wrong, the second Gram-Schmidt problem on the homework I got nothing right and on the Test I got v1 and v2 unchanged and v3 turned into [2/3,1/3,1/3,1/3,2/3,1/3] I was wondering if that was correct and what I’m doing wrong.

Again you forgot to normalize.  See my solution below

****************************************
Note ||x|| = √((-1)^2 + 2^2 + (-2)^2) = √9 = 3;   ||y||= √((-1)^2+1^2 + 0^2) = √2;
||z|| = √(1^2 + (-5)^2 +(-1)^2) = √27

SO u1 = x/||x|| = | -1/3 |
                           | 2/3  |
                           | -2/3 |
Now y - (u1Ty) u1 = | -1 |                                                        | -1/3 |
                                  |  1  |  -  ((-1/3)(-1) +(2/3)(1)+(-2/3)(0)) |  2/3 |
                                  |  0  |                                                        |-2/3  |

                                     | -1 |           | -1/3 |      | -2/3  |
                                  = |  1  |  -   1  |  2/3 |   = |  1/3  |
                                     |  0  |           |-2/3 |       | 2/3  |
and you can check that this vector is perpendicular to u1 and v1. It's also already normalized because its norm is √((-2/3)^2 + (1/3)^2 + (2/3)^2 = √(4/9 + 1/9 +4/9) = 1 so

         |-2/3|
u2 =  | 1/3|
         | 2/3|

Now we need to compute
                                              | 1 |                                                           | -1/3 | 
z - (u1Tz) u1 -(u2Tz ) u2  = | -5 |  -   ((-1/3)(1) +(2/3)(-5)+(-2/3)(-1)) |  2/3 | 
                                              |-1 |                                                          |-2/3  |
   
                                                                                                                      | -2/3 | 
                                                              -   ((-2/3)(1) +(1/3)(-5)+(2/3)(-1))  |  1/3 | 
                                                                                                                      |  2/3  |

   | 1 |               | -1/3 |           | -2/3 |
= |-5 |  -    (-3) |  2/3 |  - (-3) |  1/3 |
   |-1 |               |-2/3 |            | 2/3 |

   | 1 -  1  -  2   |       | -2  |
= |-5 -(-2)-(-1) |  =  | -2  |
   |-1 -  2 - (-2) |      | -1  |

This vector is perpendicular to u1 and u2 and also perpendicular to v1 and v2 but it's not normalized. It's norm is 3 though, so

         |-2/3|
u3 =  |-2/3|
         |-1/3|

I was also wondering if the dimension of the kernel of A was 0 because you could row reduce it to the identity matrix. 

Yes.

And is there a difference between b and g, the orthogonal basis of S and the basis of S perpendicular?

Thank you

Well yes there would have to be a difference because the subspace S and the subspace S^⟘ are perpendicular to each other, so the vectors in a basis for one would have to be perpendicular to ALL of the vectors in the other 

A practice midterm, no answers.

But you can ask specific questions about the problems. Especially if you document what you tried, and/or what you read in the textbook about the problems.

Practice Exam 2

OK, the review.

FIRST OF ALL: DON'T PANIC.

Second: Do the exam 2 practice test from the ASU mat242 website. EVERYTHING IN IT IS RELEVANT TO YOUR EXAM.  EVERYTHING IN IT HAS BEEN COVERED IN CLASS. It even has the answers.  Here is a link to it, so you don't even have to go look for it.


Third, here are some links that might be useful:

Finding matrixes, eigenvalues and eigenvectors using your calculator:
https://m.youtube.com/watch?v=n9_39MwNKGc

https://www.ottummath.com/2318/Files/Handouts/Eigenvalues%20and%20Eigenvectors%20using%20the%20TI-84.pdf

http://www.math.umbc.edu/~campbell/Calculators/TI8x/linalg.html#Matrix-Cmds

Lectures on eigenvalues and eigenvectors:

https://www.khanacademy.org/math/linear-algebra/alternate-bases/eigen-everything/v/linear-algebra-introduction-to-eigenvalues-and-eigenvectors

Fourth: A little bit later I'll post my own practice exam, but it will be similar to the exam on the ASU website.

HW question

Hello, I was just wondering why some content on the homework due tonight and Sunday has not been covered in class? Most of the problems cover topics that you have not even mentioned.

**********************************
Welp, if it will make you feel better let's have both homework sets due on Tuesday night. For the record, the kernel is another name for the null space and the image is another name for the column space, so I suggest you get started on those. And I talked about how matrixes lead to transformations by talking about how they turn lines into other lines and hence squares into quadrilaterals, where the sides of the quadrilateral are made from the columns of the matrix, which tells you everything you need to know about how to find the matrix that gives some transformation, so you probably want to get started on that too.
The one about powers of matrixes, as the hint tells you, is really just about diagonalization which I talked about on Wednesday, and is really super easy, but I'll talk about it on Monday just to make it extra clear because it's a good example.

The ones about the dot product and orthogonality and length of vectors are about the transpose of one vector times another, as talked about quite a lot and talked about the relationships between these ideas, so you might want to get started with those too.  I haven't done Gram-Schmidt algorithm yet though, so I'll do that Monday. (It's actually kind of a problematic algorithm because of numerical instability and people use other methods to get an orthogonal set instead, but it's traditional so let's do it)

6.1#8

Hello professor Taylor,

I was wondering about how I would go about finding the basis for the eigenvector for the 位1 value of -1. I understand the way you find it when the 位 is zero, as it is simply gotten by putting matrix A in RREF form and then solving for the null space of it, however, I am not sure how to work this out for the 位 value of -1.
The first thing I did was find the eigenvalues by taking the determinate of the matrix after adding in the -位 values to it. This yielded 位^4+2位^3+位^2. After solving this system of equations I got 位=0 (with multiplicity 2 because it was 位^2=0) and 位=-1 (with multiplicity 2 because it was (位+1)^2.
For the eigenspace of -1 I calculated:
-----------------
|1,   0, -1, -1|
|-1,  0,  0,  0|
|-2,  0,  2,  2|
|2,   0, -1, -1|
-----------------
Then, putting this in RREF I got
---------------
|1, 0,  0,  0|
|0, 0,  1,  1|
|0, 0,  0,  0|
|0, 0,  0,  0|
---------------
This would mean that the null space is:
-----
| 0|
| 0|
|-1|
| 1|
-----
However, this wasn't the basis which the question was asking for. Where did I mess up? Thank you in advance for your help!
**********************************
Some of your math isn't rendering properly, so I have to guess a little bit.
First of all, the null space is a space and not a vector. You could hope that the vector you wrote is a basis for the null space, but it turns out this is not the case. You haven't done completely badly however, in fact you've done everything properly except interpret the meaning of RREF(A-(-1)I).  When you solve the equation
---------------  ----      ---
|1, 0,  0,  0|    |x1|      |0|
|0, 0,  1,  1|    |x2|   = |0|
|0, 0,  0,  0|    |x3|      |0|
|0, 0,  0,  0|    |x4|      |0|
---------------  ---       ---
you get the equations x1=0,  and x3+x4 = 0. You deduce correctly that x3 must equal -x4, but you've missed the fact in all of this that x2 is not constrained at all!  Thus all vectors of the form
____            ____             _____
|  0  |            |   0   |            |   0   |
| x2 |            |   1   |            |   0   |
|-x4 |  =   x2 |   0  |   +  x4 |  -1   |
| x4 |            |   0  |             |  -1   |
-----             ------              -------
So you get two basis vectors for that eigenspace, the one you computed and in addition you need [0,1,0,0].

6.1#8

Hi Professor Taylor I've been stuck on this problem for a while now and I keep getting that the basis of the eigenvalue of -4 should be [0,1,0,0] for the matrix
|0,0,0,0|
|0,0,-4,-4|
|-4,0,2,2|
|-4,0,-2,6|
I'm not sure if I'm doing my math wrong or not. Thank you for your time.

******************************
Well, you've computed that A-(-4)Identity correctly, and you've guessed that [0,1,0,0] is in the null space. But you're also tying yourself in knots, because the eigenvectors of -4 will be all vectors in the null space of  A-(-4)Identity, which you can solve easily by doing row reduction on  A-(-4)Identity. (Hint: there's more than one vector in a basis for this null space)

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Scores and Estimated Grade

section 1.3#14

Hello Professor Taylor, I have been stuck on this problem for what seems like weeks and I can't figure it out. First, I multiplied AB and BA to get:
AB= |x^2 x^2-5y|
         |xy    xy-6y |
and
BA= |x^2-5x xy-6x |
         |-5y       -6y     |
I understand why (0,0) is a solution, but every other set of points I try don't work. My thought process in solving this is to set each similar location equal to each other and then solve a system of equations, but then the values i get only work for that one system, not the overall matrix.
Ex:
x^2 = x^2 -5x
x^2 = x(x-5)
x = 0
But, I can't think of any other possible values that might work. Could you help lead me to the right direction? Thank you very much for your time!

*******************
OK, the simple answer is that you multiplied BA incorrectly :

BA = |x   x|  |x  -5|    =   | x*x + x*y    x*(-5) + x*(-6) |   =  | x^2+xy   -11x |
         |0   y|  |y  -6|         | 0*x + y*y    0*(-5) + y*(-6) |       |    y^2       -6y  |

     =  | x^2+xy   -11x |
         |    y^2       -6y  |

This gives me the simultaneous equations

x^2 = x^2 + xy ==> xy = 0 ==> x=0 or y=0
xy = y^2 ==> y^2=0 because xy=0 from line just above hence y=0
xy = -6y ==> trivial from line just above
x^2-5y = -11x ==> x^2+11x = 0 (since y=0) hence x(x+11)=0 hence x=0 or x=-11.
Since we already know that y=0 the solutions are
(0,0) and (-11,0)

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