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Given a complex symmetric matrix $A=A^\top$ with a matrix of eigenvectors $C$ (which have distinct non zero eigenvalues) it can be shown that $C^\top C=I$ and that $C^\top A C=D$ where $D$ is a diagonal matrix of the eigenvalues.

I want to know how to show that further to the above $C^\top C= C C^\top =I$. And if this is not always true when it will be true.

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  • $\begingroup$ why don't you orthonormalize the eigen vectors?? $\endgroup$ – tattwamasi amrutam May 3 '15 at 11:36
  • $\begingroup$ I really doubt, that your statement ist true, because finding such a Matrix $C$ such that $C^T C=I$ is equivalent to being normal. But every normal matrix is diagonalizable but complex symmetric matrices aren't diagonalizable in general $\endgroup$ – Dominic Michaelis May 3 '15 at 11:43
  • $\begingroup$ I think that if $C$ and $A$ have inverses then the above must be true since: $D^2=C^\top A^2 C= C^\top A C C^\top A C$ $\endgroup$ – Joe May 3 '15 at 11:53
  • $\begingroup$ this is true as soon as the eigenvalues are simple. $\endgroup$ – mookid May 3 '15 at 12:02
  • $\begingroup$ Thanks mookid, do you have a proof or reference? $\endgroup$ – Joe May 3 '15 at 12:03
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Based on the condition that $C^\top C=I$ it follows that since $C$ is a square matrix $C C^\top=I$ see:

If $AB = I$ then $BA = I$

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  • $\begingroup$ (I'm new to stackexchange so if someone with experience would tell me if I should just delete this question that would be helpful) $\endgroup$ – Joe May 3 '15 at 16:21

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