When is the matrix of eigenvectors of a complex symmetric matrix orthogonal?

Question

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.

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 — Dominic Michaelis, May 03 '15 at 11:43
I think that if $C$ and $A$ have inverses then the above must be true since: $D^2=C^\top A^2 C= C^\top A C C^\top A C$ — Joe, May 03 '15 at 11:53
If $C^T C= I$ then $C^T $ is the inverse of $C$ and those commute. hence $C C^T=I$ too — Dominic Michaelis, May 03 '15 at 12:46
@Joe FYI. Also, could you provide (at least) a reference on the fact you claim to be true? — Algebraic Pavel, May 03 '15 at 13:52
Thanks yes see my reply to the first answer. (please let me know if I have done something very silly) — Joe, May 03 '15 at 14:02

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

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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 $BA = I$

edited Apr 13 '17 at 12:20

Community

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answered May 03 '15 at 16:20

Joe

43

(I'm new to stackexchange so if someone with experience would tell me if I should just delete this question that would be helpful) – Joe May 03 '15 at 16:21

When is the matrix of eigenvectors of a complex symmetric matrix orthogonal?

1 Answers1