my proof for the best rank approximation, using the Frobenius norm, is as follows: \begin{equation} \begin{alignedat}{1} \vert\vert{\mathbf{A-B}}\vert\vert_F^2 &= <\mathbf{A - B}, \mathbf{A-B}>_F \\ &= <\mathbf{A, A}>_F - <\mathbf{A, B}>_F - <\mathbf{B, A}>_F + <\mathbf{B, B}>_F \\ &= \sum_{i = 1}^{q} \sigma_i^2(\mathbf{A}) - 2~\text{real}~\text{trace} \left\{\mathbf{AB^H}\right\} + \sum_{i = 1}^{q} \sigma_i^2(\mathbf{B}) \\ &\geq \sum_{i = 1}^{q} \sigma_i^2(\mathbf{A}) - 2 \sum_{i = 1}^{q} \sigma_i(\mathbf{A})\sigma_i(\mathbf{B}) + \sum_{i = 1}^{q} \sigma_i^2(\mathbf{A}) \\ &= \sum_{i = 1}^{q} (\sigma_i(\mathbf{A}) - \sigma_i(\mathbf{B}))^2 \\ &= \sum_{i = 1}^{k} (\sigma_i(\mathbf{A}) - \sigma_i(\mathbf{B}))^2 + \sum_{i = k+1}^{q} \sigma_i^2(\mathbf{A}) \\ &\geq \sum_{i = k+1}^{q} \sigma_i^2(\mathbf{A}), \end{alignedat} \end{equation} where $\mathbf{B}$ is a rank $k$ matrix, $\mathbf{A}$ is rank $q$ (full rank) and has distinct singular values, $<\cdot, \cdot>_F$ denotes the Frobenius inner product and the singular values are sorted decreasingly.
Clearly the last inequality is tight if and only if the $k$ non-zero singular values of $\mathbf{B}$ are equal to the $k$ largest of $\mathbf{A}$. But does this apply to the first inequality as well (von Neumann's trace theorem)? Basically I'm trying to find the necessary and sufficient conditions (if and only if) to make the above an equality.
I'd appreciate any help!