Proving ($\left|\left|Ax\right|\right| = \left|\left|x\right|\right|$, for all $x\in\mathbb{C}^n$) $\implies A$ is unitary As the title states, I'm trying to prove that $\left|\left|Ax\right|\right| = \left|\left|x\right|\right|$ for all $x\in\mathbb{C}^n\implies$ $A$ is unitary, where $A$ is a square matrix.
This is part of a larger problem and the only thing I 'know' for the purposes of this proof is that $\left|\left|Ax\right|\right| = \left|\left|x\right|\right|$ for all $x\in\mathbb{C}^n$, nothing else, e.g. I don't know that the rows or columns of $A$ are orthonormal.
Any hints on where I can start with this? I'm quite stuck.
Thank you very much for your time.
 A: By Polarization identities (see http://en.wikipedia.org/wiki/Polarization_identity), we have:
$$(Ax,Ay)=(x,y),\forall x,y\in \mathbb{C}^n,$$
which implies your result immediately.
A: This proof is by induction. The base case is $n=2$.
We have, from the statement about the norms, that $x^{*}x=x^{*}A^{*}Ax$ $\forall x\in \mathbb{C}^2$. In particular, it is true for column vectors $e_1$ and $e_2$, the standard basis of $\mathbb{C}^2$.  Plugging these in, we can see that the $(1,1)$ and $(2,2)$ entries of $A^{*}A$ are $1$. We can also see that the $(1,2)$ and $(2,1)$ entries of $A^{*}A$ are conjugates of each other, say $u$ and $\bar{u}$.
Next let $x=[1, 1]^T$. This gives us $2=2+u+\bar{u}$. Similarly, $x=[1, i]^T$ gives $2=2+u-\bar{u}$. From these equations we can conclude that both the real and imaginary parts of $u$ are $0$, so $u=0$. Hence $A^{*}A=I_2$
Now suppose $n>2$ and let $U=A^{*}A$. Let $x \in \mathbb{C}^n$ such that all components are $0$ except the $i$th and $j$th. Given that $x^{*}x=x^{*}Ux$ we have that the principal submatrix $U[{i,j}]=I_2$. From this we conclude that $U=I_n$.
