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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?

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    $\begingroup$ you have the same upper bound on trace, given by von-neumann trace inequality. that's about all you can say. $\endgroup$ Mar 22 at 17:51
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No, the statement is not true. As a counterexample, consider $$ A = Q = V = \pmatrix{1&0\\0&1}, \quad U = \pmatrix{0&1\\1&0}. $$

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