Norm of an operator and its matrix Let $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be linear map, and define 
$$\| T\|_1= \sup \{ |T(x)|:x\in \mathbb{R}^n, |x|=1\}.$$
Let $(e_1,e_2,\cdots, e_n)$ be the standard basis of $\mathbb{R}^n$, and $[T]=[t_{ij}]$ denote the matrix of $T$ w.r.t. it. Define 
$$\| T\|_2 = \max \{|t_{ij}| : 1\le i,j\le n\}.$$
Are these two norms same? (I know that topologically all norms on finite dimensional normed space are equivalent.)
 A: In general,
$\Vert T \Vert_1 \ne \Vert T \Vert_2, \tag 0$
as the following example illustrates:
Let
$T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end {bmatrix}; \tag 1$
let
$\vec x = (x_1, x_2)^T; \tag 2$
note that
$\Vert T \Vert_2 = 1; \tag 3$
however, 
$T \vec x = (x_1 + x _2, x_2)^T; \tag 4$
thus
$\Vert T \vec x \Vert^2 = (x_1 + x_2)^2 + x_2^2; \tag 5$
taking
$x_1 = x_2 = \dfrac{1}{\sqrt 2} \tag 6$
yields
$\Vert \vec x \Vert^2 = x_1^2 + x_2^2 = 1, \tag 7$
so
$\Vert \vec x \Vert = 1, \tag 8$
but
$\Vert T \vec x \Vert^2 = (\dfrac{2}{\sqrt 2})^2 + (\dfrac{1}{\sqrt 2})^2 = \dfrac{5}{2}, \tag 9$
so
$\Vert T \vec x \Vert_1> \sqrt{\dfrac{5}{2}} > \sqrt 2; \tag{10}$
since $\Vert \vec x\Vert = 1$, this implies
$\Vert T \Vert_1 > \sqrt 2; \tag{11}$
we see that
$\Vert T \Vert_2 \ne \Vert T \Vert_1. \tag{12}$
In the above, I have taken 
$\Vert \vec x \Vert = \sqrt{x_1^2 + x_2^2}, \tag{13}$
which is probably the most widely used norm on $\Bbb R^2$.  Whether the result (12) binds for other norms on $\Bbb R^2$ I do not address here.  Nor do I have at my immediate disposal an inequality which I am sure holds for all $T$.
