Proving that $\|A\|_{\infty}$ the largest row sum of absolute value of matrix $A$ I am studying matrix norms. I have read that $\|A\|_{\infty}$ is  the largest row sum of absolute value and $\|A\|_{1}$ is the highest column sum of absolute values of the matrix $A$. However, I am not able to prove this. Are there any proof of these statements? Please help  and thanks for your time.
Edit: Where $\|A\|_{\infty}$ is the matrix norm induced by the vector norm $\|x\|_{\infty}$.
 A: I assume that the author tries to derive the matrix norms $\|A\|_1$ and $\|A\|_\infty$ induced by vector norms $\|x\|_1$ and $\|x\|_\infty$.
Let's take, for example, $\|\cdot\|_\infty$. We write
$$\|Ax\|_\infty=\sup_i \left|\sum_j A_{ij}x_j\right|\le \sup_i  \sum_j |A_{ij}||x_j|\le \sup_j |x_j| \sup_i  \sum_j |A_{ij}|.$$
Hence, a good candidate for $\|A\|_\infty$ is $\sup_i  \sum_j |A_{ij}|$. We need to prove that this boundary is indeed achieved; it is true, since we can take $i$ where that supremum is achieved and impose $x_j=\mathrm{sign}(A_{ij})$. With such $x$ all our inequalities degenerate to equalities and we conclude that $\|A\|_\infty = \sup_i  \sum_j |A_{ij}|$ is a matrix norm induced by $\|x\|_\infty = \sup_j |x_j|$.
The case $\|\cdot\|_1$ is done likewise.
A: I will give you some guidelines then. Suppose an $m$-by-$n$ matrix $A = (a_{ij})$ is given. For any $n$-dimensional vector $x$,
\begin{align*}
\|Ax\|_1 & = \sum_{i=1}^m \left|\sum_{j=1}^n a_{ij}x_j\right| \\
& \le \sum_{i=1}^m \sum_{j=1}^n \left|a_{ij}x_j\right| \\
& = \sum_{j=1}^n \left|x_j\right| \sum_{i=1}^m \left|a_{ij}\right| \\
& = \sum_{j=1}^n \left|x_j\right| A_j
\end{align*}
where I define $A_j = \sum_{i=1}^m \left|a_{ij}\right|$. If $J = \operatorname{argmax}_j A_j$, i.e., $A_J$ is a maximum among all $A_j$'s, then
$$
\|Ax\|_1 \le A_J \sum_{j=1}^n |x_j| = A_J \|x\|_1.
$$
This shows that $\|A\|_1 \le A_J$. Next, you show that there exists $x$ such that $\|Ax\|_1 = A_J\|x\|_1$. This is quite simple as you can choose $x_i = 0$ for all $i \ne J$, and $x_J = 1$ or $-1$.
For the case of $\|\cdot\|_\infty$, the situation is quite similar. (Choosing $x$ in the last step might be a bit tricky as you need to pick the right signs.) I will leave that to you.
