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?

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…
