# Prove operator norm cannot be longer than len(basis) times max(norm(basis))operator norm

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 $$v \in V , ||Tv|| ≤ mn$$.

• Welcome to MSE. What are your thoughts on the problem?
– MSDG
Dec 8, 2018 at 23:25
• What is $A$? Does $A = T$? Dec 8, 2018 at 23:28
• I was confused about that too, may be a typo from the professor, let's treat A=T Dec 8, 2018 at 23:29
• 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? Dec 8, 2018 at 23:34
• I think you might need $\Vert Tv \Vert \le mn \Vert v \Vert$. since $\Vert Tv \Vert$ must depend on $v$. Dec 8, 2018 at 23:39

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 $$||T||\leq m\sqrt{n}.$$