For a $m \times n$ matrix $A$, how to show $\Sigma_{k=1}^{r}\sigma_k(A) = max\{trace(U^T A V)\}$, where $U$ is m by r, V is n by r, and $U^TU = V^TV = I$? ( $\sigma_k(A)$ are the singular values of A )
From singular value decomposition, $A$ can be decomposed to some $U_2 \Sigma V_2^T$, however, the $U_2$ and $V_2$ are not necessarily same as the U and V before, both in terms of value and dimension. So how to prove that statement?