Eigenvalues and eigenvectors of a matrix

[blob84]

Hello i have this matrix $\in Z$ mod $7$,
M = \begin{pmatrix} 0&6\\ 5&0 \end{pmatrix}
always modulo $7$ in $Z$.
I found characteristic polynomial $x^2+5$.
Eigenvalues are $\lambda = 3, \lambda' = 4$
Eigenvectors related to $\lambda = 3$ are the non-zero solution of the system:
$4x +6y = 0,$
$5x+4y = 0$
I get:
$4x = y,$
$6y$
I don't know if it is correct, but how can i find the eigenvectors?

 

[micromass]

Yes, y=4x is correct. Now you just need to find a non-zero vector (0,0) that satisfies this. So, we can take y=1 for example. In that case (x,y)=(4,1) satisfies the equation. So (4,1) is an eigenvector.
We could also take y=2, then y=(1,2) was an eigenvector. etc.

 

[blob84]

now i have a problem to solve this:
in $Z$ mod $11$

M = \begin{pmatrix} 0&2\\ 7&0 \end{pmatrix}
The characteristic polinomyal is : $x^2+8$,
eigenvalues $\lambda=5, \lambda'=6$,
eigenvectors related to $\lambda=5$ are the non-zero solution of the system:
$6x+2y=0$
$7x+6y=0$
I don't know how to solve this system because i took a look at the solution of the exercise and it is:
$V={(y, 8y)}$
I don't know how to get this solution.
Oh i get now I'm wrong to write the first equation.

 

[HallsofIvy]

Yes, y=4x is correct. Now you just need to find a non-zero vector (0,0) that satisfies this. So, we can take y=1 for example. In that case (x,y)=(4,1) satisfies the equation. So (4,1) is an eigenvector.
We could also take y=2, then y=(1,2) was an eigenvector. etc.

You have this backwards. If y= 4x, then the eigenvector is (1, 4), not (4, 1).