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Can complex matrices with complex eigenvalues have real eigenvectors?

Say I have a complex matrix A with complex eigenvalues $\lambda_i$, under what conditions of A can the corresponding eigenvectors $\vec{V_i}$ all be real (if any)?

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    $\begingroup$ Sure: the matrix $A=\begin{bmatrix}i&0\\0&i\end{bmatrix}$ has every vector in $\mathbb{C}^2$ as an eigenvector, so in particular the vectors with real entries. $\endgroup$ – carmichael561 Jun 5 '17 at 18:25
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    $\begingroup$ One can force this to be true by starting with a diagonalized version, then multiplying it back to a valid matrix. Say we wish for eigenvectors $\{ \mathbf{v}_i \}$ to have corresponding eigenvalues $\{ \lambda_i \}$. Such a matrix $A$ can be written as $A = PDP^{-1}$, where $P$ has $\mathbf{v}_k$ as its $k^\text{th}$ column vector and $D$ is a diagonal matrix with $\lambda_k$ as the entry in its $(k,k)$ position. Now just select the eigenvalues to be complex and the vectors real-valued. $\endgroup$ – Kaj Hansen Jun 5 '17 at 18:29
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    $\begingroup$ @StuartBarth Keep in mind that the eigenvector of a given eigenspace -- even a 1D eigenspace -- is not unique. Any complex multiple of the eigenvector is also an eigenvector in that eigenspace. E.g. if you have a matrix with all complex eigenvalues and one of the eigenvectors is something like $\pmatrix{ 1 \\ 0}$ then $\pmatrix{2+2i \\ 0}$ is also an eigenvector in that eigenspace. $\endgroup$ – user137731 Jun 5 '17 at 18:32
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    $\begingroup$ Good point @Bye_World $\endgroup$ – Kaj Hansen Jun 5 '17 at 18:33

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