0
$\begingroup$

Consider a spectral decomposition of a unitary matrix $U$ given by $WAW^*$ where $A$ is diagonal matrix of eigen-values of $U$ and the symbol $^*$ means transconjugate. An infinitesimal shift $dU$would change the matrices by $dA$ and $dW$. Can anyone please help me to prove numerically that $WdW^*+dWW^*=0$. Any comments would be highly appreciated. I can see how Hermitian property can lead to the result but I need help to show numerically. Specifically, is there a way to compute $dW$ without any model of $W$.

$\endgroup$
0
$\begingroup$

Since $U$ is normal, you can choose $W$ as a unitary matrix s.t. $U=WAW^*$ and $A$ is diagonal complex. Since $WW^*=I$, we deduce that $WdW^*+dWW^*=0$ (with the transconjugate $W^*$ and not with the transpose $W'$ !).

$\endgroup$
5
  • $\begingroup$ I see that a brave people downvoted my post without leaving his name... $\endgroup$ – user91684 Apr 9 '16 at 12:38
  • $\begingroup$ @ Creator .The important thing is not the downvote, but that 1. your question is absolutely incomprehensible. 2. There is a big mistake that you do not want to correct. Thus, completely rewrite your question or I'll vote to close it. $\endgroup$ – user91684 Apr 11 '16 at 14:19
  • $\begingroup$ May I ask what is it the big mistake that I do not want to correct? If I know the mistake, I will correct it or delete the question myself, thanks. $\endgroup$ – Creator Apr 11 '16 at 18:02
  • $\begingroup$ @ Cretor . I just see that $W'$ became the transconjugate !! You really are the only one using this notation. $\endgroup$ – user91684 Apr 12 '16 at 2:21
  • $\begingroup$ I did not know how to use * got it from you now. thanks. I can delete the question but since it has an answer it complains. $\endgroup$ – Creator Apr 12 '16 at 3:45

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.