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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?

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  • $\begingroup$ Look at the excellent answer here math.stackexchange.com/questions/235396/eigenvalues-are-unique $\endgroup$ – fixedp Jul 28 '16 at 10:58
  • $\begingroup$ Please, write the characteristic polynomial as a product of irreducible factors, so that it is obvious what the eigenvalues are. Btw, the same eigenvalue appearing twice is perfectly valid, it would just have algebraic multiplicity greater than 1. $\endgroup$ – lisyarus Jul 28 '16 at 10:59
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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 $\;n\;$.

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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 $1$.

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