Minimization of matrix $2$-norm

Question

I have following optimization problem
\begin{equation}
\underset{B\in \mathbb{R}^{nd}}{\text{min }} \|I_n-BA^T\|_2
\end{equation}
where $d<n$, $A\in\mathbb{R}^{nd}$ and full column rank. It is equivalent to
\begin{equation}
\underset{B\in \mathbb{R}^{nd}}{\text{min }}\underset{\|x\|_2^2=1}{\text{max }}\|x-BA^Tx\|^2_2
\end{equation}
Is it on the right way to change the order of min and max?
i.e.
\begin{equation}
\underset{\|x\|_2^2=1}{\text{max }}\underset{B\in \mathbb{R}^{nd}}{\text{min }}\|x-BA^Tx\|^2_2
\end{equation}
The KKT conditions of inner optimization implies
\begin{equation}
B=xx^TA(A^Txx^TA)^{-1}
\end{equation}
If it's right then I get stuck at the outer optimization. Any help would be appreciated!

Answer 1

Let $P=I-BA^T$. Since $\operatorname{rank}(BA^T)=d<n$, the product $BA^T$ is singular. Therefore $Px=x$ for some nonzero vector $x$. It follows that $\|P\|_2$ is always bounded below by $1$, and $P$ is nonzero.

This lower bound $1$ is attained by the trivial solution $B=0$, but there is a solution with full column rank. Let $B=A(A^TA)^{-1}$. As $A$ has full column rank, $B$ is well defined and has full column rank. Now $P=I-A(A^TA)^{-1}A^T$ is a nonzero orthogonal projection (i.e. $P^T=P=P^2$). Hence $\|P\|_2=1$.

This $B$ ($=A(A^TA)^{-1}$) is known as the Moore-Penrose pseudo-inverse of $A$.