What happens to the trace if you multiply with orthogonal matrices

Suppose we are given matrices $$A$$ and $$Q$$. Furthermore denote by $$U$$ and $$V$$ orthogonal matrices. For a matrix $$Q'$$ the equality $$Q'=UQV^T$$ holds. Since $$U$$ and $$V$$ are orthogonal matrices, the matrices $$Q$$ and $$Q'$$ have the same singular values. Denote by $$\langle A, B \rangle = \mbox{tr} (A^T B)$$ the Frobenius inner product for two matrices $$A$$ and $$B$$. A fellow student now claims that $$\langle Q, A \rangle = \langle Q', A \rangle$$, but unfortunately I do not see why. Does somebody know if this is true?

• you have the same upper bound on trace, given by von-neumann trace inequality. that's about all you can say. Mar 22 at 17:51

No, the statement is not true. As a counterexample, consider $$A = Q = V = \pmatrix{1&0\\0&1}, \quad U = \pmatrix{0&1\\1&0}.$$