Matrix that maximizes that trace problem

Question

I am trying to prove that the optimal solution for the following problem

$$\begin{array}{ll} \underset{X \in \Bbb R^{n \times m}}{\text{maximize}} & \mbox{Tr} (AX^T)\\ \text{subject to} & \|X\|_2 \leq 1\end{array}$$

is $U_r V_r^T$ where $A = U_r \Sigma_r V_r^T$ is the compact SVD form (non-zero singular values), and the maximum value is $\mbox{Tr}(\Sigma_r)$.

I think this problem can be linked to the following problem, however, I can't seem to find the connection.

$$\max_{u\in \mathbb{R}^m,v\in \mathbb{R}^n:||u||_2=1,||v||_2=1}u^TAv=\sigma_1(A)$$

Any help would be greatly appreciated.

Answer 1

Von Neumann's trace inequality states that $$ \mathrm{tr}(\mathbf{A} \mathbf{B}^T) \le \sum a_i b_i $$ where $a_1$ is the largest singular value of $\mathbf{A}$, and $b_1$ is the largest singular value of $\mathbf{B}^T$ and so on.

Because of the constraint $$ \mathrm{tr}(\mathbf{A} \mathbf{B}^T) \le \sum a_i $$ This bound is achieved by choosing $\mathbf{B}=\mathbf{U}_r \mathbf{V}_r^T$ together with the SVD $\mathbf{A}=\mathbf{U}_r \mathrm{diag}(\mathbf{a}) \mathbf{V}_r^T$ so that $\mathbf{A} \mathbf{B}^T= \mathbf{U}_r \mathrm{diag}(\mathbf{a}) \mathbf{U}_r^T$ and $$ \mathrm{tr}(\mathbf{A} \mathbf{B}^T) = \sum a_i $$ because trace remains invariant in similarity transformation.