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).