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).

 

[blue_raver22]

Hello,

Thank you for sharing your work on finding the eigenvalues and eigenvectors of the given matrix. Your approach is correct, but there are a few steps missing that will help you find the eigenvectors.

To find the eigenvectors, we need to solve the system of equations you have written for the corresponding eigenvalue. In this case, we have the system:

4x + 6y = 0
5x + 4y = 0

To solve this system, we can use any method we prefer, such as substitution or elimination. I will use substitution in this response.

First, we can rearrange the first equation to get y in terms of x:

y = -4/6 x

Now, we can substitute this value of y into the second equation:

5x + 4(-4/6x) = 0
5x - 8/3x = 0
(5 - 8/3)x = 0
x = 0 or x = 0

Since we cannot have zero as an eigenvector, we can ignore the first solution and focus on x = 0. Now, we can plug this value of x into the first equation to find the corresponding value of y:

4(0) + 6y = 0
y = 0

Therefore, the eigenvector corresponding to the eigenvalue \lambda = 3 is [0, 0]. You can follow a similar approach to find the eigenvector corresponding to the eigenvalue \lambda' = 4.

I hope this helps clarify the process of finding eigenvectors. Keep up the good work!