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Is it true that for symmetric matrix with complex entries all eigenvalues are real.

I have seen the proof for Hermitian matrices and proved it for real symmetric matrices,but for complex symmetric matrices I dont know how to determine this.

Thank you in advance for your help.

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  • $\begingroup$ Hint: Try a 2x2 matrix where all entries = $i$. $\endgroup$ – Amzoti Mar 5 '13 at 5:09
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    $\begingroup$ This is true for self-adjoint matrices, aka hermitian matrices. Not for symmetric matrices. Look at $\left(\matrix{0&i\\i&0}\right)$ for instance. $\endgroup$ – Julien Mar 5 '13 at 5:10
  • $\begingroup$ @julien: easier example than mine! Regards $\endgroup$ – Amzoti Mar 5 '13 at 5:12
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    $\begingroup$ How about $\begin{bmatrix} i \end{bmatrix}$. No simpler than that! $\endgroup$ – copper.hat Mar 5 '13 at 5:54
  • $\begingroup$ @copper.hat: Nice example! $\endgroup$ – Amzoti Mar 5 '13 at 6:06
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The question has been answered in the comments; this is a community wiki answer to allow the question to be marked as answered.

The answer is no; symmetric matrices with complex entries do not in general have real eigenvalues.

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