# eigenvalues and eigenvectors of a matrix

by blob84
Tags: eigenvalues, eigenvectors, matrix
 P: 24 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?
 Emeritus Sci Advisor Thanks PF Gold P: 15,868 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.
 P: 24 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.
Math
Emeritus