Exercise 1.8.10

Remove first the $N-n$ columns of $A$, the new matrix has $\left|\right|C\left|\right|\le \left|\right|A\left|\right|$, this is because the Frobenius matrix norm is a $l^2$ norm of a $\mathit{mn}$ vector, the less the number of elements in the matrix, the less of its norm.

Then transpose $C$: no change in norm. Finally remove $M-m$ columns of $C^T$ to product $B^T$ with no increase in norm. Altogether $\left|\right|B\left|\right|=\left|\right|B^T\left|\right|\le \left|\right|C^T\left|\right|=\left|\right|C\left|\right|\le \left|\right|A\left|\right|$.