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I am trying to show the following. If $A$ is linear on a finite dimensional inner product space. How do I show that $||Ax|| =||A^{*}x||$ for all $x$ implies that $A$ is normal?

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What's the norm? –  Martin Argerami Sep 12 '12 at 3:03
    
Sorry it is an inner product space –  Digital Gal Sep 12 '12 at 3:06
    
I think Schurs Theorem will work here. –  Digital Gal Sep 12 '12 at 3:14
    
.... since the theorem is easy to prove if the matrix representation is upper triangular –  Digital Gal Sep 12 '12 at 3:32

1 Answer 1

up vote 5 down vote accepted

If the norm is induced by the inner product, then we have $$\| Ax\|^2= \langle Ax, Ax\rangle = \langle A^*Ax, x\rangle$$ Likewise $\| A^*x\|^2=\langle AA^*x, x\rangle$

Since $\| A^*x\|^2=\| Ax\|^2$ holds for any $x$, we have $AA^*=A^*A$

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How is the last line $\| A^*x\|^2=\| Ax\|^2$ different than the hypothesis? –  Digital Gal Sep 12 '12 at 3:51
    
$\| A^*x\| = \| Ax\|$ <=> $\| A^*x\|^2=\| Ax\|^2$ –  chaohuang Sep 12 '12 at 3:59
4  
A slight elaboration is in order: @chaohuang has shown that $\langle (A^*A - A A^*) x , x \rangle = 0$ for all $x$. Since $(A^*A - A A^*)$ is self-adjoint, this implies that $A^*A - A A^* = 0$. (This property is not true in general, eg, take a rotation in $\mathbb{R}^2$.) –  copper.hat Sep 12 '12 at 3:59
    
got it thanks ! –  Digital Gal Sep 12 '12 at 4:02

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