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

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 $A^*A-I$.

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Daniel Fischer, drhab, amWhy, Michael Albanese, Moron
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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 $A^*A-I=0$.