Unitary matrix matrix problem I need to proof this question Show that for any
Recall that the unitary group $U(n)$ consists of all $A \in M_n(C)$ with $A^*A = I$. Show that a matrix $A \in M_n(C)$ is in $U(n)$ if and only if $\langle Ax,Ayi\rangle = \langle x,y\rangle$ for all $x, y \in C$.
So I just need the $\Leftarrow$ implication; I made it but I am not sure if my justification is good enough. 
$\langle Ax,Ay\rangle = \langle x,A^*Ay\rangle = \langle x,y\rangle = x^*A^*Ay = x^*y$ so {problem here} $A^*A = I$
therefore $A \in M_n(C)$ is in $U(n)$.
So do you guys think that this is good enough or do is their some problems in this argument?
 A: Hints:
1.) For $T,S\in M_n(\Bbb C)$ we have $\ T=S\ \iff\ \forall x,y:\langle Tx,y\rangle=\langle Sx,y\rangle$. 
2.) Use the adjointness property: $\langle z,Ay\rangle=\langle A^*z,y\rangle$.
A: It seems that what you need to show is that
$$
x^*(A^*A)y = x^*y \text{ for all } x,y \in \Bbb C^n \implies
A^*A = I
$$
In general: suppose that $x^*Ay = x^*By$ for all $x,y \in \Bbb C^n$.  How can we use this to conclude that $A = B$?
A: The argument seems a bit confused.
When you write
$$
\langle Ax,Ay\rangle = \langle x,A^*Ay\rangle = \langle x,y\rangle = x^*A^*Ay = x^*y
$$
the last equality is just a rephrasing of $\langle Ax,Ay\rangle = \langle x,y\rangle$, so it gives nothing new.

The hypothesis is that $\langle Ax,Ay\rangle = \langle x,y\rangle$ for all $x,y$ and you want to show that $A^*A=I$.
For all $x,y$ we have
$$
\langle x,y\rangle=\langle Ax,Ay\rangle=\langle x,A^*Ay\rangle
$$
which means that
$$
\langle x,y-A^*Ay\rangle=0
$$
But, if $v$ is a vector such that $\langle x,v\rangle=0$ for all $x$, then $v=0$, so, $y-A^*Ay=0$ for all $y$. Hence…
