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Math Help - Prove This Is A Norm

  1. #1
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    Prove This Is A Norm

    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.
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  2. #2
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    Quote Originally Posted by mathematicalbagpiper View Post
    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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