Question

Find the eigenvalues and the corresponding eigenvectors of the stated power of A = [1 3 7 11; 0 0.5 3 8; 0 0 0 4; 0 0 0 2]; A^9

Answer 1

Since this is a diagonal matrix, the eigenvalues are the values on the diagonal.
I'll write out a solution and post it, it might take abit though, this is a lot of work (at least for me it is lol)

Answer 2

Well I found that it was a triangular matrix so I got the eigenvalues but can't get the eigenvectors

Answer 3

just to make sure I have this right, the matrix is:
1 3 7 11
0 .5 3 8
0 0 0 4
0 0 0 2
right?

Answer 4

Yeah so since it's lower triangular aren't the values just 1, .5 2 and 0?

Answer 5

yeah, thats right, so now we have to go through the process of figuring out the eigenvectors >.<

Answer 6

Could you explain how to do one because I don't even know where to start

Answer 7

Actually I do but keep getting it wrong

Answer 8

like if it's 0 would the matrix be a zero matrix subtract matrix A

Answer 9

you need to create the matrix:
\[A - \lambda I\]
and see what its null space (or kernel, whatever terminology you use) is.

Answer 10

Is it not the other way around λI-A?

Answer 11

either or, it depends on how you were taught. I was taught the other way, but Ive seen it the way you have it in other books.

Answer 12

I think the \[A-\lambda I\] is much easier to deal with. It doesn't matter once you multiply things out and get the character equations.

Answer 13

okay then I row reduce it then I don't know where to go from there I got [0 1 0 0; 0 0 1 0; 0 0 0 1; 0 0 0 0]

Answer 14

is that A row reduced?

Answer 15

yes this is for when λ = 1

Answer 16

Alright, so now you have the matrix:
0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
This means that the x_1 variable is free, and that all the other variables must be 0

Answer 17

remember your working with:
\[(\lambda I - A)x = 0\]

so your solutions are in the form:
\[x = x_1(1, 0, 0, 0)\]
the vector (1, 0, 0, 0) spans the null space with the eigenvalue 1, so it is its corresponding eigenvector.

Answer 18

Yeah that's the thing I don't get, how you get it into that form?

Answer 19

like the book first shows x1 = t then you take t out so t(1 0 0 0) or something

Answer 20

Now you have to do the same process with the other eigenvalues to get the other 3 eigenvectors. Ah, one sec I write out how to get it in that form, its easier to show that it is to say it lol

Answer 21

Alright thanks

Answer 22

Alright, basically if the column doesnt have a pivot, then its a free variable. You can write all the vectors that are in the null space of a matrix in terms of those free variables.

Answer 23

That make sense thanks! So now I got all the eigenvectors for the four eigenvalues, what do I do now? How about the A^9

Answer 24

would the eigenvectors for the eigenvalues of A^9 be the same as they are proportional?

Answer 25

i'll write it out, one sec lol >.<

Answer 26

Haha sorry

Answer 27

Haha omg got it

Answer 28

Yeah it is the same which make sense! thanks for all your help!

Answer 29

alright, your goal is to create
\[A = P\Lambda P^{-1}\]
Where Lambda is a matrix formed by oh you got it? sweet sweet

Answer 30

no wait

Answer 31

sry for taking so long, i wanted to make sure I explained it properly lol

Answer 32

yours so much more complex??

Answer 33

I thought you find the eigenvectors for the matrix A then find the eigenvectors for the matrix A^9??

Answer 34

and they turn out to be the same values?

Answer 35

except different eigenvalues

Answer 36

oh i though we were trying to calculate A^9 >.<

Answer 37

haha actually the eigenvalues are the same except ^9 ohhh I got it now

Answer 38

Ohh haha okay well you explained a lot to me either way thanks for your help!

Answer 39

yeah you got it, thats right

Answer 40

lolol my bad my bad, i was making a mountain out of a mole hill >.< its because i like Linear Algebra =/