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Let $V=\mathbb{C}$ or $\mathbb{R}$, $||\cdot||_{\infty}$ a norm on our vector space and $A:V\to V$ a linear transformation. We define $$||A||_M := \sup_{x\in V \\ ||x|| \leq 1} ||Ax||_{\infty} $$ Assuming $(A_{ij})$ is the matrix $A$ in standard basis, prove that: $$||A||_M = \max_{1 \leq i\leq n} \sum_{j=1}^n |A_{ij}|$$

It feels very "logical" but somehow I have no idea how I go about writing a rigorous proof. Thank you for your time.

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  • $\begingroup$ There are two steps: first, show that $\|Ax\|_\infty$ is necessarily less than the number on the right if $\|x\|_\infty\leq 1$. Then, show that there exists an $x$ so that $\|x\|_\infty=1$ and $\|Ax\|_\infty$ is equal to the number on the right. $\endgroup$ – Omnomnomnom Jan 16 '18 at 14:37
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Define $\bar{A}=\max_{1 \leq i\leq n} \sum_{j=1}^n |A_{ij}|$.

For $x=(x_1, \dots, x_n)$ with $\Vert x \Vert_\infty \le 1$ and $1 \le i \le n$ you have $$\left\vert \sum_{j=1}^n A_{ij}x_j \right\vert \le \sum_{j=1}^n \vert A_{ij}x_j \vert \le \Vert x \Vert_\infty \left(\sum_{j=1}^n |A_{ij}| \right) \le \sum_{j=1}^n |A_{ij}| \le \bar{A}.$$

Hence $\Vert A \Vert_M \le \bar{A}$.

Now take $i_0 \in \{1, \dots, n\}$ such that $\sum_{j=1}^n |A_{i_0 j}| = \bar{A}$ and $x_0 = (\text{sign}(A_{i_0 1}), \dots, \text{sign}(A_{i_0 n}))$.

You have $\Vert x_0 \Vert_\infty = 1$ and $\Vert A x_0 \Vert = \bar{A}$ providing the conclusion.

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