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Can a matrix of complex numbers have real eigenvalues ? Imaginary part of the complex number are not equal to zero.

I know Hermitian matrix can have real eigenvalues, but what about non-Hermitian ones?

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  • $\begingroup$ if this weren't possible, quantum mechanics would be in trouble! (in other words, see the spectral theorem.) $\endgroup$ – symplectomorphic May 1 '14 at 17:30
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Take for instance the matrix $$ m =\begin{bmatrix} 1&0\\ 0&2 \end{bmatrix} $$ and conjugate it with a complex matrix; for instance, conjugate $m$ with $$ t = \begin{bmatrix} 1+i & i\\2+i & i\end{bmatrix} $$ to get $$ t^{-1} m t = \begin{bmatrix}3+i & i \\ -3+i & -i \end{bmatrix}, $$ which has still eigenvalues $1$ and $2$.

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