I wasn't $100\%$ sure in the accuracy of what I obtained, so may I ask for verification?

Let $A,B\in M_n(\Bbb C)$ be two similar Hermitian matrices. Then $A=P^{-1}BP$.

According to the spectral theorem, every Hermitian matrix is diagonalizable, so $A=U_1^{-1}DU_1$ and $B=U_2^{-1}DU_2$, where $U_1,U_2\in M_n(\Bbb C)$ are unitary and $D=\left(\delta_{ij}\right)\in M_n(\Bbb C)$ is diagonal s. t. $\delta_{ii}\in\sigma(A)=\sigma(B)$.

Since $A$ and $B$ are unitarily similar to $D$, I wanted to write $A$ in the following form: $$A=P^{-1}U_2^{-1}DU_2P$$ Then $U_2P=U_1\implies P=U_2^{-1}U_1$. Since both $U_1$ and $U_2^{-1}$ are unitary, $P$, as a product of two unitary matrices, is also unitary.

Question according to this result:

Are all similar Hermitian matrices unitarily similar?

If this is valid, can it be used in the proof that the matrix representation of a Hermitian operator in an orthonormal basis is a Hermitian matrix? The statement is obvious when, given an arbitrary Hermitian matrix, one is asked to diagonalize it. I thought $P$ could be the transition matrix from one orthonormal basis to another one.

Thank you in advance!

  • $\begingroup$ I am having a hard time following what you are saying. It is true that every Hermitian matrix is unitarily equivalent to a diagonal matrix. But two diagonal matrices are unitarily equivalent if and only if they have the same diagonal elements, up to writing them in a different order. So the diagonal matrix whose diagonal entries are 1, 1, and 2 is not unitarily similar to the diagonal matrix whose diagonal entries are 2, 2, and 1. $\endgroup$ Aug 17 '20 at 20:22
  • $\begingroup$ @StephenMontgomery-Smith, I didn't mean unitarily equivalent, but unitarily similar. I'm wondering if Hermitian matrices are similar, are they unitarily similar? Let's say we take your example of a diagonal matrix with entries $\operatorname{diag}\{1,1,2\}$. Then, I mean, $A$ is unitarily similar to $D$ and $B$ is unitarily similar to $D$. Then, my problem boils to the possible transitivity of the relation "being unitarily similar". I hope it sounds less messy. We used the term equivalent only for the matrices of the same type and rank to distinguish the similarity. $\endgroup$
    – Invisible
    Aug 17 '20 at 20:31
  • 1
    $\begingroup$ Sorry. I thought that similar and equivalent meant the same thing in this context. $\endgroup$ Aug 17 '20 at 20:34
  • 1
    $\begingroup$ Anyway, is your question this? "If two Hermitian matrices are similar, then they are unitarily similar?" If that is the question, then the answer is "yes." $\endgroup$ Aug 17 '20 at 20:34
  • $\begingroup$ @StephenMontgomery-Smith, thank you for verifying! And more importantly, for the patience! $\endgroup$
    – Invisible
    Aug 17 '20 at 20:35

As verifed by @StephenMontgomery-Smith, all similar Hermitian matrices are unitarily similar, indeed.

Howerver, $P$ doesn't necessarily have to be unitary if, e. g., $A=B=0$. Then every invertible $P$ would suffice.

You may see this parallel question I posted on Quora, answered yesterday by Aaron Dunbrack.


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.