Norm preservation properties of a unitary matrix Tried to prove the following facts:


*

*If $B \in M_{n\times n}(\mathbb{C})$ is unitary  (i.e. $ B^{-1} = B^{*}$), then:


*

*$\forall x\in \mathbb{C}^n :\| Bx \|_2 = \| x \|_2  $

*$\forall A\in M_{n\times n}(\mathbb{C})$ :  $ \| BA \|_F = \| A \|_F  $ (where $ \| . \|_F $ denotes Frobenius norm).


*$\forall A,B\in M_{n\times n}(\mathbb{C}): \| AB \|_{ \infty} \leq \| A \|_{ \infty}\| B \|_{ \infty}   $
In (1),tried to conduct direct algebraic manipulations from the definitions of the norms, but I obtained no results. 
  In (2) the use of definition $ \| B \|_{ \infty} := max\{\frac{\|Bx\|}{\|x\|}: x\in \mathbb{C}^n \setminus \{0\} \}$ and the inequality $\|AB\| \leq \|A\|\|B\|$ gives straightforward result. However, the use of definition $\|B\|\ _{\infty} :=max\{ \sum_{j=1}^{n}|b_{ij}|: i\in \{1,...n\}\}$ does not give direct results.
I would be thankful for hints/advices!
 A: The trick here is to operate with the squares of the norms, rather than with the norms themselves.  These squares are given by the appropriate scalar products, whose properties we will use.
Using the Hermitian product, $(x, y)$:
$$
|| Bx ||_{2}^{2} = (Bx, Bx) = (x, B^{*}Bx) = (x, x) = ||x||_{2}^{2}.
$$
For the Frobenius norm, using the corresponding inner product:
$$
||B A||_{F}^{2} = \mbox{tr }[(B\,A)^{*} B\,A] = \mbox{tr }[A^{*} B^{*} B \, A] = \mbox{tr }[A^{*} \, A] = ||A||_{F}^{2}
$$
A: $$
||Bx||_2 = \sum_i\left(\sum_j B_{ij}x_j) \right)^* \left(\sum_k B_{ik}x_k) \right) $$
$$
\\
\left(\sum_j B_{ij}x_j) \right)^* = \sum_j B_{ij}^*x_j^*  =  \sum_j (B^T_{ji})^*x_j^* =  \sum_j B^{-1}_{ji}x_j^* 
$$
$$
||Bx||_2 = \sum_i\left(\sum_j B^{-1}_{ji}x_j^*  \right)^* \left(\sum_k B_{ik}x_k) \right) \\
= \sum_j\sum_k \left(  \sum_iB^{-1}_{ji}  B_{ik}  \right) x_j^*x_k
=\sum_j\sum_k \delta_{jk} x_j^*x_k = \sum_j  x_j^*x_k = ||x||_2
$$
The equality for the Frobenius norm follows directly since for any arbitrary $x$ the sum of the quares for each column of $BAx$ matches the sum of the squares for that same column of $Ax$.
