If $A$ and $B$ are unitarily equivalent, then $\sum_{i,j}|A_{ij}|^2 = \sum_{i,j}|B_{ij}|^2$

Question

If $A$ and $B$ are unitarily equivalent, then $\sum_{i,j}|A_{ij}|^2 = \sum_{i,j}|B_{ij}|^2$

We have a unitary matrix $U$ such that $U^*AU = B$

How can I proceed from here to show $\sum_{i,j}|A_{ij}|^2 = \sum_{i,j}|B_{ij}|^2$?

Is this correct?

$BB^*=U^*AUU^*A^*U=U^*AA^*U$

so $\text{tr}(BB^*)=\text{tr}(AA^*)$

hence

$\text{tr}(AA^*)=\sum_{i,j}|A_{ij}|^2 = \sum_{i,j}|B_{ij}|^2 =\text{tr}(BB^*)$