Prove the induced matrix norm $\|A\|_\infty = \max_i \| a^*_i \|_1$

Question

For $A \in \mathbb{C}^{m \times n}$, prove the induced matrix norm $$\|A\|_\infty = \max_i \|a^*_i\|_1$$ where $a^*_i$ denotes the $i$th row of $A$.

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Attempt:

proof

Using the $a_j$ to denote the columns of $A$ we have

$$\|A\|_\infty := \sup_{\|x\|_\infty =1}\|Ax\|_\infty = \sup \left|\left| \sum_{j=1}^n x_ja_j\right|\right|_\infty \leq \sup \sum_{j=1}^n \left|\left|x_ja_j \right|\right|_\infty \leq \sup \sum_{j=1}^n |x_j| \|a_j\|_\infty \cdots$$

Here's were I'm having some difficulty

$$ \cdots\leq \sup \left\{ \max_j ||a_j||_\infty \sum_{j=1}^n |x_j|\right\} = \sup \left\{ \left(\max_j \|a_j\|_\infty\right) \|x_j\|_1 \right\} \leq \sup \left\{ \left(\max_j \left[\max_{1\leq i \leq m}a_{ij} \right] \right) \|x_j\|_1 \right\} \cdots$$

I think I'm doing something fatally wrong here but I can't figure it out. Style wise this also seems like way too much.

Answer 1

You have $$ \left\|\sum_j x_ja_j\right\|_\infty =\max_k\left|\sum_j a_{kj}x_j \right| \leq\max_k\sum_j|a_{kj}|\,|x_j| \leq\max_k\sum_j|a_{kj}|=\max_k\|a_k^*\|_1. $$ And then you need to show that the equality is achieved, which is easily seen to be the case when $x=e$, the vector with all entries equal to $1$.