# Thread: Prove This Is A Norm

1. ## Prove This Is A Norm

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.

2. Originally Posted by mathematicalbagpiper
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.
$\sup\{ |\langle (A+B)u,u \rangle|:||u||=1 \}=\sup\{ |\langle Au+Bu,u \rangle|:||u||=1 \}$ $=\sup\{ |\langle Au,u \rangle +\langle Bu,u\rangle|:||u||=1 \}$ $\leq \sup\{ |\langle Au,u \rangle|:||u||=1 \}+\sup\{ |\langle Bu,u \rangle|:||u||=1 \}$ ,
since $|\alpha+\beta|\leq |\alpha|+|\beta|$ .