# Unitary equivalence of matrices with equal trace

It is obvious that $B = OAO^\dagger$ have equal trace if $O$ is Hermian. But is is also so that:

$Tr(A) = Tr(B) \rightarrow A = OBO^\dagger$

For some Hermitian $O$ ?

I know this should be trivial to prove/disprove but Is just can't get it sorted out...

• Ps: sorry for the bad Tex. Stackexchange doen't deel to like me phone's dollar sign :/ – gertian Jun 28 '17 at 11:48
• What do you mean by $O^{\dagger}?$ – RideTheWavelet Jun 28 '17 at 12:09
• The hermitian conjugate – gertian Jun 28 '17 at 12:35
• I think you mean that $O$ is unitary (and maybe that $A$ and $B$ are Hermitian) – Omnomnomnom Jun 28 '17 at 12:40

I think that you meant that $O$ is unitary, and that $A$ and $B$ are Hermitian. If that's the case, your statement is not true.
Note that $OAO^\dagger$ has the same eigenvalues as $A$. If we take $$A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{2&0\\0&0}$$ Then we see that $\operatorname{tr}(A) = \operatorname{tr}(B)$. Since they have different eigenvalues, however, there can be no unitary $O$ such that $B = OAO^\dagger$.
I am that sure that I understood your question, but I suppose that $A=\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ is a counter-example.