0
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ Hint: Try a 2x2 matrix where all entries = $i$. $\endgroup$ – Amzoti Mar 5 '13 at 5:09
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.