How come two of the eigenvalues are same? Question is about finding the eigenvalues of the matrix :
$$\begin{bmatrix}
0 & 0 & 2 \\
0 & 2 & 0 \\
2 & 0 & 0 \\
\end{bmatrix}$$
the matrix would become
$$\begin{bmatrix}
-A & 0 & 2 \\
0 & 2-A & 0 \\
2 & 0 & -A \\
\end{bmatrix}$$
$$
-A(-A(2-A))-2(2)(2-A))=0
$$
from here how come two eigenvalues appear? I didn't get that, how come two same numbers could hold two eigenvalues at the same time?
And btw if the matrix is 3x3 are there always 3 eigenvalues ? Is this the same case for a nxn matrix?
 A: Why would you use $\;A\;$ instead of $\;x,t,\lambda\;$ or some other more or less standard notation I can't say, but notation many times helps to make things clearer...or messier, of course.
Let us calculate the characteristic polynomial of the given matrix $\;A\;$ :
$$\det(tI-A)=\begin{vmatrix}t&0&-2\\0&t-2&0\\-2&0&t\end{vmatrix}=t^2(t-2)-4(t-2)=$$$${}$$
$$=(t-2)(t^2-4)=(t-2)^2(t+2)$$
We thus have the double eigenvalue $\;t=2\;$ and the simple one $\;t=-2\;$
Any square $\;n\times n\;$ matrix has at most $\;n\;$ different eigenvalues, but as this case shows the number of different eigenvalues can be less than $\;n\;$.
A: A different  approach would be the following: 
Let 
$$B=\begin{vmatrix}0&0&1\\0&1&0\\1&0&0\end{vmatrix}$$
Since $A$  is a  multiple of $B$  all facts about eigenvalues are the same. 
Observe that  $B$ is a matrix which permutes the first and third component of a vector $x$ and leaves the middle component unchanged. 
In this case we observe that a   possible eigenvalue is $\lambda_1=1$ and the two corresponding independent   eigenvectors  are $e_1=(1,0,1)^{'}$  and  $e_2=(0,1,0)^{'}$.
The multiplicity of $\lambda_1$ is $2$.
The second eigenvalue is   $\lambda_2=-1$ with basis eigenvector $e_3=(-1,0,1)^{'}$ and multiplicity  $1$.
