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In this problem using the characteristics equation it comes out that there are four complex roots two identical each . So according to question there is no eigenvalue on $\mathbb{R}$ but in $\mathbb{C}$. Is my explanation justified?

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    Yes, there are no real eigenvalues, so you are correct. – 2017-01-20
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    You could even just focus on each of the blocks, since the two zero blocks mean that they operate separately on their own planes. That makes it easier to see that the only possible eigenvalues are $\pm i$. – 2017-01-20

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Yes, you're right. Since your matrix is of the style

$$\left(\begin{matrix} A & 0\\ 0 & -A\\ \end{matrix}\right) $$

Then your characteristic polynomial of that matrix is the same as the polynomial of $A$ squared multiplied by $-1$ (That's the reason of why you have repeated roots). The characteristic polynomial of $A$ is $$(1-\lambda)(-1-\lambda)-(\sqrt{2})(-\sqrt{2})=\lambda^2-1+2=\lambda^2+1$$

Then you have that the characteristic polynomial is $-(\lambda^2+1)^2=0$, then the only factor is $\lambda^2+1$, so equaling to zero you have that $\lambda^2=-1$, so we can say that the roots are $\pm i$.

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    Monsieur Galois according to your characteristics polynomial lambda is zero . isn't it. I think some mistake is in it. Correct me if I am wrong – 2017-01-20
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    Let me a moment to check it. – 2017-01-20
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    Why you say that the characteristic polynomial says that $\lambda=0$? – 2017-01-20
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    If we put -1+2=(lambda)^2+1it gives lambda ^2 +1 =1 which gives lambda equals 0. – 2017-01-20
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    What I did was expand the factors of the characteristic polynomial. That is equal to $\lambda^2-1+2$, that is equal to ^$\lambda^2+1$, so then, equaling to zero, you will have that $\lambda^2+1=0$. – 2017-01-20
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    You have written it right here but above if u see its still lambda square + one = one. But I got it . thanks – 2017-01-20
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    Could you point me which line are you saying? – 2017-01-20
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    Just above -( Then you have the characteristic polynomial is -(lambda^2+1)^2, and can.... – 2017-01-20
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    I think it was clear before, but I added more explanation to make it helpful for you. – 2017-01-20