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The conclusion I got was A has no real eigenvalues.

Can anyone tell me if it's right or what are these eigenvalues?

juliana
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    You're right: the eigenvalues are not in $\mathbb R$. However, $A$ does have complex eigenvalues; see greedoid's answer. It is actually often the case that a real $n\times n$ matrix will have complex eigenvalues. – Dave Dec 02 '18 at 20:30

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If $Av = \lambda v$ then $$ A^2v = A( \lambda v) = \lambda (Av) = \lambda^2v $$

But $$A^2v = -Iv=-v \implies ( \lambda^2+1)v = 0$$

So $ \lambda =\pm i$.

nonuser
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  • I do this. If $A\in\mathbb{R}^{n\times n}$ there is no problem of the eigenvalues being in $\mathbb{C}$? – juliana Dec 02 '18 at 20:22
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    No, ther is no problem with $\mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots. – nonuser Dec 02 '18 at 20:32
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    You can see something similary here https://math.stackexchange.com/questions/2314703/find-eigenvalues-of-a-whena-is-3-times3-real-valued-matrix-such-that-a3-i?rq=1 – nonuser Dec 02 '18 at 20:33
  • Yes, there is a problem. Look at the case $n=3$. –  Dec 02 '18 at 21:47