If the tensor product of $A$ and $B$ is Hermitian, are $A$ and $B$ Hermitian? Given $A \otimes B$ is Hermitian, does it follow that $A$ and $B$ are necessarily Hermitian?
I've only gotten as far as $(A \otimes B)^\dagger = A^\dagger \otimes B^\dagger = A \otimes B$ and I haven't been able to prove that $A = A^\dagger$ and $B = B^\dagger$.
 A: No, this is not true. If $A$ and $B$ are anti-Hermitian, that is if:
$$A^\dagger = -A, B^\dagger = -B$$
then their tensor product is Hermitian:
$$(A\otimes B)^\dagger = (-1)^2 A\otimes B$$
A: No. Counterexample: $A=0$ and $B$ is anything.
A: The way to answer this question is to think in terms of a basis for the matrix, for convenience we can choose a basis that is hermitian, so for a 2-by-2 matrix it has basis:
$$ e_{1} = \left(\begin{matrix} 1 & 0 \\ 0 & 0\end{matrix} \right), \qquad e_{2} = \left(\begin{matrix} 0 & 0 \\ 0 & 1\end{matrix} \right), \qquad e_{3} = \left(\begin{matrix} 0 & 1 \\ 1 & 0\end{matrix} \right), \qquad e_{4} = \left(\begin{matrix} 0 & -i \\ i & 0\end{matrix} \right)$$
now we can think about matrices as elements of a vector space:
$$M = \alpha e_{1} + \beta e_{2} + \gamma e_{3} + \delta e_{4}$$
If $\alpha, \beta, \gamma, \delta$ are complex numbers than we can write any arbitrary matrix, however if we restrict them to be only real numbers then we instead can write any Hermitian matrix.
This approach extends to any n-by-n matrix (using an $n^2$-dim basis). If we now consider matrix $A$ as having basis $e_{i}$ and coefficients $\alpha_{i}$ while consider matrix $B$ as having basis $h_{j}$ and coefficients $\beta_{j}$. The constraint that $(A\otimes B)^{\dagger} = A\otimes B = \sum_{i,j} \alpha_{i}\beta_{j} e_{i}\otimes h_{j}$ implies that
$$\alpha_{i}\beta_{j} \in \mathbb{R} \quad \forall \ i,j$$
What this means is that if we can write $\alpha_{1} = |\alpha_{1}| e^{i\theta}$ (or any non-zero coefficient) then we know all $\beta_{j} = |\beta_{j}| e^{-i\theta}$, and therefore all $\alpha_{i} = |\alpha_{i}| e^{i\theta}$ (note this includes zeros).
Therefore you can't say that $A$ and $B$ are hermitian but rather you can say that there is a $\theta$ so that both $A e^{-i\theta}$ and $B e^{i\theta}$ are Hermitian.
This excludes an edge case where either $A$ or $B$ are zero and so my proof above requires the additional assumption that $A \neq 0$ and $B\neq 0$, explaining the counter example given by Javier in their answer. Furthermore this also shows that the general counter example given by jacob1729 in their answer is the only kind ofcounter example to the original statement (for non-trivial $A$ and $B$).
