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.

  • $\begingroup$ 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

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 $BA = I$

| cite | improve this answer | |
  • $\begingroup$ (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

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.