Is the spectrum of a product of two operators, $AB$, invariant under $UAU^{\dagger}$ for unitary $U$?

Question

This question is about linear operators on a Hilbert space. If necessary, the Hilbert space can be assumed to be finite dimensional. I have two Hermitian operators, $A$ and $B$. Do we have

$$ \sigma(AB) = \sigma(UAU^{\dagger}B), $$

for a unitary $U$? $\sigma(X)$ denotes the spectrum of $X$.

Answer 1

This not generally the case.

Finite dimensional counter-example: $$ \{0,1\} = \sigma\left[\pmatrix{0&1\\0&1} \right] =\sigma\left[\pmatrix{1&1\\1&1}\pmatrix{0&0\\0&1} \right] \neq\\ \sigma\left[\pmatrix{0&0\\0&2}\pmatrix{0&0\\0&1} \right] =\sigma\left[\pmatrix{0&0\\0&2} \right] = \{0,2\} $$ Note in particular that $$ \pmatrix{1&1\\1&1} = U\pmatrix{0&0\\0&2}U^* $$ with $U = \frac{1}{\sqrt{2}}\pmatrix{1&1\\-1&1}$