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Can a real eigenvalue (of a matrix that has two real and two complex conjugate eigenvalues) correspond to a complex eigenvector? Thanks in advance.

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    $\begingroup$ Sure, and you don't even need the complex eigenvalues. E.g., $[i , 0]$ is an eigenvector of the identity matrix. $\endgroup$ – eyeballfrog Apr 27 at 17:54
  • $\begingroup$ @eyeballfrog though the identity matrix does not have necessarily complex eigenvectors $\endgroup$ – Henry Apr 27 at 17:56
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Yes. Take, for instance, the matrix$$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix}.$$Its eigenvalues are $1$ (multiplicity two), $i$, and $-i$. And $(i,i,0,0)$ is an eigenvector corresponding to the eigenvalue $1$.

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