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?
Normal Operator
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linear-algebra
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0What's the norm? – 2012-09-12
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0Sorry it is an inner product space – 2012-09-12
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0I think Schurs Theorem will work here. – 2012-09-12
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0.... since the theorem is easy to prove if the matrix representation is upper triangular – 2012-09-12
1 Answers
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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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0How is the last line $\| A^*x\|^2=\| Ax\|^2$ different than the hypothesis? – 2012-09-12
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0$\| A^*x\| = \| Ax\|$ <=> $\| A^*x\|^2=\| Ax\|^2$ – 2012-09-12
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5A 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 $\mathbb{R}^2$.) – 2012-09-12
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0got it thanks ! – 2012-09-12