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How to prove that an $n\times n$ matrix $A$ is unitary if and only if $\|Az\| = \|z\|$ for all $z \in \mathbb{C}^n$?

Suggestion: Apply previous question on "How to prove that $T$ is the $0$ operator, that is $T(v) = 0$ for all $u,v \in V$ ?" to $A^*A-I$.

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closed as off-topic by Daniel Fischer, drhab, amWhy, Michael Albanese, Moron Jul 30 at 12:05

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This is standard theory, found in many texts. –  Robin Chapman Nov 5 '10 at 10:21
-1 : just because the OP isn't making any effort at all and keep spamming with his homework questions. –  Djaian Nov 5 '10 at 10:36

1 Answer 1

up vote 2 down vote accepted

I suppose that your definition of unitary is that $A^*A=AA^*=I$.
Let $A$ be unitary and $x\in V$. Then $$\|Ax\|^2=\langle Ax,Ax\rangle=\langle x, A^*Ax\rangle =\langle x,Ix\rangle=\langle x, x\rangle=\|x\|^2.$$

Conversely, suppose $\|Ax\|=\|x\|$ for all $x\in V$. Then for all $x,y\in V$, one has $$\|A(x-y)\|^2=\|Ax-Ay\|^2=\|Ax\|^2-2\langle Ax,Ay\rangle+\|Ay\|^2=\|x\|^2-2\langle Ax,Ay\rangle+\|y\|^2$$ and $$ \|x-y\|^2=\|x\|^2-2\langle x,y\rangle+\|y\|^2.$$ Equating $\|A(x-y)\|^2$ and $\|x-y\|^2$ yields $\langle Ax,Ay\rangle=\langle x,y\rangle$. Thus, for all $x,y$, $$\langle x,(A^*A-I)y\rangle=\langle x,A^*Ay\rangle-\langle x,y\rangle=\langle Ax,Ay\rangle-\langle x,y\rangle=0$$ Then it follows from your cited previous question that $A^*A-I=0$.

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