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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$\begingroup$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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