Prove This Is A Norm

**mathematicalbagpiper** · March 29th 2010, 07:33 AM

Problem reads as follows:

Let V be a finite real vector space of dimension n that admits an orthonormal basis $\{u_1,....u_n\}$ . Then the set $\{ |\langle Au,u \rangle|:||u|| = 1 \}$ is bounded. Hence the set has a supremum. Let $||A||=sup\{ |\langle Au,u \rangle|:||u||=1 \}$ . Then it can be argued from analysis that there exists a vector $u_0$ such that $||A|| = \langle Au_0, u_0 \rangle$ . Prove that $||A||$ defines a norm.

I've already proven everything but the triangle inequality. Here's where I'm getting stuck.

**tonio** · March 29th 2010, 07:58 AM

Originally Posted by mathematicalbagpiper

Problem reads as follows:

Let V be a finite real vector space of dimension n that admits an orthonormal basis $\{u_1,....u_n\}$ . Then the set $\{ |\langle Au,u \rangle|:||u|| = 1 \}$ is bounded. Hence the set has a supremum. Let $||A||=sup\{ |\langle Au,u \rangle|:||u||=1 \}$ . Then it can be argued from analysis that there exists a vector $u_0$ such that $||A|| = \langle Au_0, u_0 \rangle$ . Prove that $||A||$ defines a norm.

I've already proven everything but the triangle inequality. Here's where I'm getting stuck.

$\sup\{ |\langle (A+B)u,u \rangle|:||u||=1 \}=\sup\{ |\langle Au+Bu,u \rangle|:||u||=1 \}$ $=\sup\{ |\langle Au,u \rangle +\langle Bu,u\rangle|:||u||=1 \}$ $\leq \sup\{ |\langle Au,u \rangle|:||u||=1 \}+\sup\{ |\langle Bu,u \rangle|:||u||=1 \}$ ,

since $|\alpha+\beta|\leq |\alpha|+|\beta|$ .

Tonio

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