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

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.

• why don't you orthonormalize the eigen vectors?? – tattwamasi amrutam May 3 '15 at 11:36
• 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 3 '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 3 '15 at 11:53
• this is true as soon as the eigenvalues are simple. – mookid May 3 '15 at 12:02
• Thanks mookid, do you have a proof or reference? – Joe May 3 '15 at 12:03

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 $BA = I$

• (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 3 '15 at 16:21