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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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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0got it thanks ! – 2012-09-12