Proof that $\|Ax\|_2=\|x\|_2$ if $A$ is an unitary matrix and $ A \in C^{m \times m}$ How to prove that $$\|Ax\|_2=\|x\|_2$$
 if $A$ is an unitary matrix and $ A \in C^{m \times m}$
I know that the multiplication is a change of basis in the coordinate system. But how can I get a formal proof?
 A: Noting
$$ \|x\|_2^2=\bar x^Tx$$
one has
$$ \|Ax\|_2^2=(\bar A\bar x)^TAx=(\bar x^T\bar A^T)Ax=\bar x^T(\bar A^TA)x=\bar x^Tx=\|x\|_2^2 $$
or 
$$ \|A\|_2=\|x\|. $$
Here
$$ \bar A^TA=I $$
is used since $A$ is a unitary matrix.
A: Hint:


*

*If $\{e_1,\dots,e_m\}$ denotes the standard basis of $\Bbb C^m$, then $\{Ae_1,\dots,Ae_m\}$ is also an orthonormal basis of $\Bbb C^m$.

*For any orthonormal basis $\{v_1,\dots,v_m\}$, and coefficients $a_1,\dots,a_m$, we have $\|a_1 v_1 + \cdots + a_m v_m\|^2 = |a_1|^2 + \cdots + |a_m|^2$.



Regarding the second point: let $A^*$ denote the conjugate transpose of $A$.  For column vectors $x,y$, $x^*y$ is the standard sesquilinear "dot-product" $\langle y,x \rangle$, and $\|x\|^2 = x^*x$.  With that, note first that.
$$
\left\| \sum_{j=1}^m a_j v_j\right\|^2 = 
\left( \sum_{j=1}^m a_j v_j\right)^*
\left( \sum_{k=1}^m a_k v_k\right) = 
\sum_{j,k = 1}^m \bar a_j a_k v_j^*v_k.
$$
By orthonormality, we have $v_j^*v_k = 0$ when $j \neq k$ and $v_j^*v_j = 1$. So, this sum is simply
$$
\left\| \sum_{j=1}^m a_j v_j\right\|^2 = 
\sum_{j=1}^m \bar a_j a_j v_j^*v_j = \sum_{j=1}^m |a_j|^2.
$$
