If $V$ is right-orthogonal, does it hold $\langle AV,BV\rangle_F=\langle A,B\rangle_F$?

Question

Let $A,B\in\mathbb R^{m\times n}$. It's easy to see that for the Frobenius inner product it holds $$\langle A,B\rangle_F=\operatorname{tr}B^\ast A=\operatorname{tr}A^\ast B.\tag1$$ So, if $U\in\mathbb R^{k\times m}$ is left-orthogonal (i.e. $U^TU=I_m$), then $$\langle UA,UB\rangle_F=\operatorname{tr}B^TU^TUA=\langle A,B\rangle_F\tag2.$$ Can we show an equivalent result for right-orthogonal (i.e. $VV^T=I_n$) $V\in\mathbb R^{n\times k}$ and $\langle AV,BV\rangle_F$?

Answer 1

This follows from the cyclic property of the trace (see Is trace invariant under cyclic permutation with rectangular matrices? ) $$ \begin{aligned} \langle AV, BV \rangle_F &= \mathrm{tr} V^T B^T A V \\ &= \mathrm{tr} B^T A V V^T \\ &= \langle A, B \rangle_F \end{aligned} $$