Inequality for subordinate norm of a submatrix : $\Vert B\Vert\le\Vert A\Vert$.

Question

We 'did' this exercise in class

Show that the subordinate norm of a submatrix is smaller than the the subordinate norm of the matrix i.e. $$\Vert B\Vert\le\Vert A\Vert$$ where $B$ is the sub matrix of $A$.

The solution was :

The main idea is to write $B=M_1AM_2$ with $\Vert M_h\Vert\le1$ for $h=\{1,2\}$ with it easy to conclude. We only said $M_1,M_2$ is the matrices with entries $1$ on the rows and columns of $A$ contained in $B$ and $0$ elsewhere but we did not wrote specifically the matrix $M_1$ and $M_2$.

Question 1: Can we write *pecifically the matrix $M_1$ and $M_2$?

Question 2: We used the inequality for subordinate norm the matrices must be square matrices, Is it possible to deal with non-square matrices?

Answer 1

Let $I$ be the set of rows chosen to make $B$ from $A$ and let $J$ be the set of columns chosen to make $B$ from $A$. We define $M_1,M_2$ by $$ M_1[i,j] = \begin{cases} 1 & i = j \in I \\ 0 & \text{otherwise} \end{cases}\\ M_2[i,j] = \begin{cases} 1 & i = j \in J \\ 0 & \text{otherwise} \end{cases} $$