Show that $\displaystyle |\|A|\|_{\infty}=\sup \limits_{1 \le i \le m} \sum_{j=1}^{n}|a_{ij}|$ Show that $$|\|A|\|_{\infty}=\sup_{\|x\|_{\infty}=1}\|Ax\|_{\infty}$$ for $A=[a_{ij}]$ an $m \times n$ matrix.
Show that $$ \||A\||_{\infty}=\sup_{1 \le i \le m} \sum_{j=1}^{n}|a_{ij}|$$
Since the sum is happening over columns , I am unable to find an element. I mean had it been over the rows I would have chosen the normal basis elements $e_i$. 
Any hint would do. Thanks for the help!!
 A: Let $A = (a_{i,j})_{1 \leq i,j \leq n}$ be a $n \times n$ real matrix. For $x = \begin{bmatrix} x_{1} & \ldots & x_{n} \end{bmatrix}^{\top} \in \mathbb{R}^{n}$, you have : 
$$
\begin{align*}
\vert (Ax)_{i} \vert &\leq {} \sum_{j=1}^{n} \vert a_{i,j} \vert \vert x_{j} \vert  \\[2mm]
 &\leq \Big( \sum_{j=1}^{n} \vert a_{i,j} \vert \Big) \Vert x \Vert_{\infty}
\end{align*}
$$
By taking the max in this inequality, you get :
$$ \Vert Ax \Vert_{\infty} \leq \max \limits_{1 \leq i \leq n} \Big( \sum_{j=1}^{n} \vert a_{i,j} \vert \Big) \Vert x \Vert_{\infty} \tag{1} $$
From $(1)$, you get : $\displaystyle |||A||| \leq \max \limits_{1 \leq i \leq n} \sum_{j=1}^{n} \vert a_{i,j} \vert$. In order to obtain $\displaystyle |||A||| \geq \max \limits_{1 \leq i \leq n} \sum_{j=1}^{n} \vert a_{i,j} \vert$, you can find a vector $x_{0} \in \mathbb{R}^{n}$ such that $\Vert x_{0} \Vert_{\infty} = 1$ and $\displaystyle \Vert Ax_{0} \Vert_{\infty} = \max \limits_{1 \leq i \leq n} \sum_{j=1}^{n} \vert a_{i,j} \vert$. For that, let $k \in \left\{ 1, \ldots, n \right\}$ such that :
$$ \sum_{j=1}^{n} \vert a_{k,j} \vert = \max \limits_{1 \leq i \leq n} \sum_{j=1}^{n} \vert a_{i,j} \vert $$
And take :
$$ x_{0} = \begin{bmatrix} \mathrm{sign}(a_{k,1}) \\ \vdots \\ \mathrm{sign}(a_{k,n}) \end{bmatrix} $$
Clearly, $\Vert x_{0} \Vert_{\infty} = 1$ and you can prove that $\displaystyle \Vert Ax_{0} \Vert_{\infty} = \sum_{j=1}^{n} \vert a_{k,j} \vert = \max \limits_{1 \leq i \leq n} \sum_{j=1}^{n} \vert a_{i,j} \vert$.
As a conclusion :
$$ |||A||| = \max \limits_{1 \leq i \leq n} \sum_{j=1}^{n} \vert a_{i,j} \vert. $$
A: Hint: the infinity norm is not relevant.
If $\|\cdot\|$ is a norm on $V$ and, for a bounded operator $T\colon V\to V$ we define
$$
|||T|||=\sup_{x\ne0}\frac{\|Tx\|}{\|x\|}
$$
then
$$
|||T|||=\sup_{\|x\|=1}\|Tx\|
$$
Once you have proved this, you have the first part. Now we can specialize to the infinity norm.
Consider the vectors $[\pm1\ \pm1\ \dots\ \pm1]^T$, which have norm $1$.
