find V such as $VAV^\dagger=UA$

Question

Given a hermitian matrix $A$ and a unitary $V$. There exists a unitary $U=VAV^\dagger A^{-1}$ such as $VAV^\dagger=UA$ and $UU^\dagger=VAV^\dagger A^{-1} AVA^{-1}V^\dagger=1$. Is the converse true ? i.e given $A$ hermitian and $U$ unitary, can we find a unitary $V$ such as $$VAV^\dagger=UA \quad?$$

Answer 1

Even if $A$ is invertible and one can write down $U=VAV^*A^{-1}$ as you did, this matrix in general is not unitary. While any generic pick for $A$ and $V$ will probably show this, an explicit example (where $A$ even is invertible) would be $$ A=\begin{pmatrix}2&0\\0&1\end{pmatrix}\quad\text{ and }\quad V=\begin{pmatrix}0&1\\1&0\end{pmatrix}\,. $$ Then $$ U=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}\frac12&0\\0&1\end{pmatrix}=\begin{pmatrix}\frac12&0\\0&2\end{pmatrix} $$ is the unique matrix which satisfies $VAV^*=UA$ but $U$ is obviously not unitary. The mistake in your calculation to check for unitarity of $U$ happens when you compute $$ U^*=(VAV^*A^{-1})^*=(A^{-1})^*VA^*V^*= \boxed{ A^{-1}}VAV^* $$ because then $UU^*=VAV^*A^{-1}A^{-1}VAV^*=VAV^*A^{-2}VAV^*$, and $A^{-2}=(A^2)^{-1}$ in general cannot be simplified further.

Similarly one sees that the statement you are actually asking about is false: again take $$ A=\begin{pmatrix}2&0\\0&1\end{pmatrix}\quad\text{ and }\quad U=\begin{pmatrix}0&1\\1&0\end{pmatrix}\,. $$ Then we are looking for a unitary matrix $V$ such that $$ V\begin{pmatrix}2&0\\0&1\end{pmatrix}V^*=\begin{pmatrix}0&1\\2&0\end{pmatrix}\,.\tag{1} $$ There are numerous ways to see that such a unitary cannot exist, but the easiest is probably taking the trace of eq. (1): because the trace is cyclic one finds $$ \operatorname{tr}(VAV^*)=\operatorname{tr}(AV^*V)=\operatorname{tr}(A)=3\neq 0=\operatorname{tr}\begin{pmatrix}0&1\\2&0\end{pmatrix} =\operatorname{tr}(UA) $$ which is the contradiction we were looking for.