Consider $\mathbb{K}^n$, $\mathbb{K}^m$, both with the $||.||_1$-norm, where $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$.

Let $||T|| = inf\{M ≥ 0: ||T(x)|| ≤ M ||x|| \space \forall x \in \mathbb{K}^n\}$ be the operator norm of a linear transformation $T: \mathbb{K}^n \to \mathbb{K}^m$.

I then want to show that the operator norm of the linear transformation $T$ is also given by:

$$||T|| = max\{\sum_{i = 1}^m|a_{ij} |, 1 ≤ j ≤ n\} =: ||A||_1$$

where $A$ is the transformation matrix of $T$ and $a_{ij}$ it's entry in the $i$-th row and $j$-the column.

Thanks in advance. I'm not very familiar with operator norms yet, so any help would be appreciated. I found this (Operator Norm of a Linear Transformation) thread that deals with a related topic, but it didn't answer my problem so far.


Note that for any $x\in \mathbb{K}^n$ $$ \|T(x)\|_1 = \sum_{i=1}^m |T(x)_i| \leq \sum_{i=1}^m\sum_{j=1}^n |a_{i,j}||x_j| \qquad (\ast) $$ For $1\leq j\leq n$, set $$ \alpha_j = \sum_{i=1}^m |a_{i,j}| $$ Then rearrange $(\ast)$ to get $$ \|T(x)\|_1 \leq \sum_{j=1}^n \alpha_j |x_j| \leq \|A\|_1 \|x\|_1 $$ This proves that $\|T\| \leq \|A\|_1$.

For the reverse inequality assume first that $a_{i,j} \geq 0$ for all $i,j$ , and suppose $$ \alpha_1 = \sum_{i=1}^m a_{i,1} = \|A\|_1 $$ Then for $x = (1,0,0,\ldots, 0)$ we have $\|x\|_1 = 1$ and $$ T(x) = (a_{1,1}, a_{2,1}, \ldots, a_{m,1}) \Rightarrow \|T(x)\|_1 = \|A\|_1 $$ and this prove that $\|T\| \geq \|A\|_1$

Now can you complete the argument for the general case?


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