Question

how do you prove that eigenvalues and eigenvectors are correct for a given matrix?

Answer 1

I have eigenvalues -1,1,2,4 with eigenvectors (0,0,1,0),(-2,2,1,0),(0,-2,-1,1), and (0,0,1,5) respectively

Answer 2

the initial matrix is \[\left[\begin{matrix}1 & 0&0&0 \\ 1 & 2&0&0\\1&2&-1&1\\1&1&0&4\end{matrix}\right]\]

Answer 3

is there any way to multiply them or something to obtain the initial matrix?

Answer 4

you need to show that\[A\vec b=\lambda\vec b\]so once you have an eigenvector \(\vec\eta_n\) you want to show that after that vector undergoes the transformation due to \(A\) it will point in the same direction, i.e. be a constant multiple of itself.

Answer 5

so\[A\vec\eta_i=\lambda_i\eta_i\]notice the change in magnitude is by a factor \(\lambda_i\), which is the \(i^{th}\) eigenvalue

Answer 6

\[A\vec\eta_i=\lambda_i\vec\eta_i\]

Answer 7

so it'll be vector=vector, right?

Answer 8

A* eigenvector = constant * eigenvector

Answer 9

you are showing that multiplying the eigenvector by the matrix is the same as multiplying the eigenvector by its eigenvalue

Answer 10

for instance if my eigenvalue \(\lambda\) is 2, and my eigenvector \(\vec\eta\) is (1,-1), then you should be able to show that \[A\vec\eta=A\left[\begin{matrix}1\\-1\end{matrix}\right]=\left[\begin{matrix}2\\-2\end{matrix}\right]=2\left[\begin{matrix}1\\-1\end{matrix}\right]=\lambda\vec\eta\]

Answer 11

okay, so for the first one, I get A*(vector) = 0 but that doesn't sound right

Answer 12

I'm not sure, haven't tried the problem yet

Answer 13

or [0//0//0//0]

Answer 14

what's the eigenvalue you are looking at?

Answer 15

-1

Answer 16

let me try, I haven't done this in a bit

Answer 17

so does lambda in this one have to be an eigenvalue?

Answer 18

not sure I get you, lambda is an eigenvalue, yes

Answer 19

I just plugged the rest into my calculator quick and they were all multiples of the eigenvector, but they weren't by one of the eigenvalues I had

Answer 20

I end up getting *0, *2, *3, and *5 for lambda for the four of them

Answer 21

so the eigenvector you got corresponding to lambda=-1 is what?

Answer 22

wait, was I solving for something else? I took A times a 4x1 matrix of the eigenvector and solved for lambda.

Answer 23

yeah, but what did you get for that eigenvector

Answer 24

I got 0 for lambda

Answer 25

I think we are having a laps in communication. You got your eigenvectors as {-1,1,2,4}, correct?
then you got the eigenvector for each eigenvalue. what eigenvector did you get for the eigenvalue -1 ?

Answer 26

sorry, the vector for -1 is (0,0,1,0)

Answer 27

good that is what I got. now multiply that by A

Answer 28

I got 0

Answer 29

I have a sneaking suspicion that you are multiplying your eigenvector \(\vec\eta\) by \(A-\lambda\) and not \(A\)

Answer 30

\[\left[\begin{matrix}1 & 0&0&0 \\ 1 & 2&0&0\\1&2&-1&1\\1&1&0&4\end{matrix}\right]\left[\begin{matrix}0\\0\\1\\0\end{matrix}\right]=?\]

Answer 31

[0,0,-1,0]?

Answer 32

yes

Answer 33

which is\[-1\cdot\left[\begin{matrix}0\\0\\1\\0\end{matrix}\right]=\lambda\vec\eta\]

Answer 34

okay. Then that would be a multiple of -1, which is the eigenvalue. I wonder what I did wrong...

Answer 35

well it's right now, you see that, correct?

Answer 36

ya I think so

Answer 37

\[A\vec\eta=\left[\begin{matrix}1 & 0&0&0 \\ 1 & 2&0&0\\1&2&-1&1\\1&1&0&4\end{matrix}\right]\left[\begin{matrix}0\\0\\1\\0\end{matrix}\right]=\left[\begin{matrix}0\\0\\-1\\0\end{matrix}\right]=-1\cdot\left[\begin{matrix}0\\0\\1\\0\end{matrix}\right]=\lambda\vec\eta~~~~~~~~~~~~\huge\checkmark\]

Answer 38

my guess is that you multiplied by \(A-\lambda\) and got\[(A-\lambda I)\vec\eta=\left[\begin{matrix}2 & 0&0&0 \\ 1 & 3&0&0\\1&2&0&1\\1&1&0&5\end{matrix}\right]\left[\begin{matrix}0\\0\\1\\0\end{matrix}\right]=\left[\begin{matrix}0\\0\\0\\0\end{matrix}\right]\]

Answer 39

yep. I didn't realize that for some reason. Thanks for the help. I think I got it from here

Answer 40

great, glad I could help :)