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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.)

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    $\begingroup$ By same, do you mean literally the same norms, or equivalent norms? $\endgroup$ – Mark Nov 12 '17 at 4:33
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    $\begingroup$ I means, is $\|T\|_1=\|T\|_2$? If not, is there any inequality which holds for all $T$? $\endgroup$ – Beginner Nov 12 '17 at 4:35
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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 $T$.

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