2
$\begingroup$

One of my professors mentioned that since a matrix A is positive semi definite and B is hermitian, hence the inner product $<A,B>$ is real. Is this an if and only if condition? So if we know that B is hermitian and $<A,B>$ is real then does it imply that A is hermitian as well?

Update 1: From the answer it seems like my answer is no. Could you also then explain, if we are given a linear map T which takes hermitian operators to hermitian operators, then why is the adjoint map hermitian as well?

$\endgroup$
0
$\begingroup$

No. Take any matrix A, and define the matrix $$\hat{A} = \overline{\langle A,B\rangle } A$$

Then

$$\langle \hat{A}, B \rangle = \langle \overline{\langle A,B\rangle } A, B\rangle = \overline{\langle A,B\rangle }\langle A, B\rangle \in \Bbb R$$

And $\hat{A}$ is generally not hermitian

$\endgroup$
  • $\begingroup$ Thanks. If you do not mind, do you have an answer to the updated question as well? Thanks since my intial goal was to solve this updated question through my earlier question which you showed to be false. $\endgroup$ – user1_1 Dec 17 '15 at 22:10

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.