1. ## Eigenvectors

Hi, I'm having trouble finding the right values...

Consider the following system I have:

2x - y + z = x
0x + 2y +0z = y
x + 3y +2z = z

So we see from our second equation that 2y=y, thus y=0, thats fine so far

However, I now find z = -x

so this gives us an eigenvector:
1
0
-1

Can someone show me how I get the 1st and 3rd entry for this vector? Because from what i know i thought the 1st entry would also be -1?

A clear explanation would be good thanks

2. For what eigenvalue? The matrix
$\begin{bmatrix}2 & -1 & 1 \\ 0 & 2 & 0 \\ 1 & 3 & 2\end{bmatrix}$
has three eigenvalues: 1, 2, and 3.

If <x, y, z> is an eigenvector, with eigenvalue 1, we must have
$\begin{bmatrix}2 & -1 & 1 \\ 0 & 2 & 0 \\ 1 & 3 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$
which gives the equations 2x- y+ z= x, 2y= y, and x+ 3y+ 2z= z. The equation 2y= y gives y= 0, of course, so 2x+ z=x or z= -x and x+ 2z= z also gives z= -x. Any vector of the form <x, 0, -x>= x<1, 0, -1> is an eigenvalue for eigenvector 1. That's probably the one you were referring to.

If <x, y, z> is an eigenvector, with eigenvalue 2, we must have
$\begin{bmatrix}2 & -1 & 1 \\ 0 & 2 & 0 \\ 1 & 3 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}$
which gives the equations 2x- y+ z= 2x, 2y= 2y, and x+ 3y+ 2z= 2z. The first equation is the same as -y+ z= 0 or z= -y and so the third is x+ 3y= 0 or x= -3y. That tells us that any eigenvector, corresponding to eigenvalue 2, is of the form <-3y, y, -y>= y<-3, 1, -1>.

If <x, y, z> is an eigenvector with eigenvalue 3, we must have
$\begin{bmatrix}2 & -1 & 1 \\ 0 & 2 & 0 \\ 1 & 3 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}3x \\ 3y \\ 3z\end{bmatrix}$
which gives the equations 2x- y+ z= 3x, 2y= 3y, and x+ 3y+ 2z= 3z. The second equation gives y= 0. Then the first equation is 2x+ z= 3z or x= z. The third equation is x+ 2z= 3z or again x= z. Any eigenvector, corresponding to eigenvalue 3, is of the form <x, 0, x>= x<1, 0, 1>.