$L: \Bbb{R}^n \rightarrow \Bbb{R}^m$ be the linear mapping then $||L|| = $ max$_{1 \leq i \leq m, 1 \leq j \leq n}$ $|a_{ij}|$.

Question

Let $L: \Bbb{R}^n \rightarrow \Bbb{R}^m$ be the linear mapping with the matrix $(a_{ij})$ with $1-$norm on $\Bbb{R}^n$ and sup-norm on $\Bbb{R}^m$.

I am trying to show that the corresponding norm on $L_{m,n}$ is $||L|| = $ max$_{1 \leq i \leq m, 1 \leq j \leq n}$ $|a_{ij}|$.

This seems to be the variation of operator norm? but how to prove this?

Answer 1

Let $M=\max \{|a_{ij}|:1\leq i \leq m, 1\leq j \leq n\}$. We have to show that $\max \{|\sum_j a_{ij}x_j|:\sum_j |x_j | \leq 1, 1\leq i \leq m \}=M$. First note that $|\sum_j a_{ij}x_j|\leq M $ if $\sum_j |x_j | \leq 1$. Given any $i_0,j_0$ consider the vector $x$ defined by $x_j=s(a_{i_0j_0})$ for $j=j_0$ and $0$ for all other values of $j$, where $s(t)=1$ if $t \geq0$ and $s(t)=-1$ if $t <0$. Then $\sum_j a_{i_0j}x_j =|a_{i_0j_0}|$ and $\|x\|=1$. Hence $\max \{|\sum_j a_{i_0j}x_j|:\sum_j |x_j | \leq 1\}\geq |a_{i_0j_0}|$ for all $i_0$ and $j_0$ . To complete the proof take maximum over $i_0$ and $j_0$.