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Let $V$ be a finite dimensional normed linear space and let $T \in \mathscr L (V)$. Define the operator norm of $T$ to be the smallest number $M$ such that $||T v|| ≤ M||v||$ for any $v \in V$ . We will write $||T||$ to mean that smallest number $M$, the operator norm.

Let $B = e_1, ..., e_n$ be an orthonormal basis for V, a normed linear space of dimension $n$. Let $T \in \mathscr L (V )$. Let $m = Max\{||T e_1||, ||T e_2||, ..., ||T e_n||\}$. That is, $m$ is the length of the longest vector in the list $T e_1, ..., T e_n$. Prove that for any vector $v \in V , ||Tv|| ≤ mn$.

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  • $\begingroup$ Welcome to MSE. What are your thoughts on the problem? $\endgroup$
    – MSDG
    Dec 8, 2018 at 23:25
  • $\begingroup$ What is $A$? Does $A = T$? $\endgroup$ Dec 8, 2018 at 23:28
  • $\begingroup$ I was confused about that too, may be a typo from the professor, let's treat A=T $\endgroup$
    – jmars
    Dec 8, 2018 at 23:29
  • $\begingroup$ As far as my thoughts go, I'm confused as to how why it's not simply $||Tv||\leq m$. Shouldn't the longest ||Tv|| be m since m is the max norm of the images of the basis? $\endgroup$
    – jmars
    Dec 8, 2018 at 23:34
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    $\begingroup$ I think you might need $\Vert Tv \Vert \le mn \Vert v \Vert$. since $\Vert Tv \Vert$ must depend on $v$. $\endgroup$ Dec 8, 2018 at 23:39

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This may be what you need. Write $v = \sum_{i=1}^n v_i e_i $ where $v_i = \langle v,e_i\rangle$. Then we have $$ ||Tv||^2 = ||\sum_{i=1}^n v_i T(e_i)||^2 \leq \sum_{i=1}^n |v_i|^2\sum_{i=1}^n ||T(e_i)||^2\leq ||v||^2nm^2, $$ by Cauchy-Schwarz inequality. Hence, it follows that $$ ||T||\leq m\sqrt{n}. $$

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