How to calculate norm of matrix using any orthogonal basis? 
Show that for $X \in \mathrm{M}_n(\mathbb{C})$ and any orthonormal basis $\{u_1, \ldots , u_n\}$
  of $\mathbb{C}^n$, we have $$\|X\|^2=\sum_{j,k}^n|\langle u_j,Xu_k\rangle |^2.$$

My Attempt:
I thought this to prove using projection formula as 
to find coordinate of $X_{ij}=\langle u_i,Xu_j\rangle$.
But I was thinking I am missing something. As this is $10$ mark question with just one line argument How it was ?
Please give me hint .I wanted to solve this problem
Any Help will be appreciated
 A: Define a unitary map by $Ue_k = u_k$ for every $k=1,2,\ldots,n$, where $\{e_1,e_2,\ldots, e_n\}$ is the standard basis of $\Bbb C^n$. Then, we have
$$\begin{align*}
\sum_{j=1}^n\sum_{k=1}^n\left|\langle u_j,Xu_k\rangle\right|^2&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle Ue_j,XUe_k\rangle\right|^2\\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle e_j,U^*XUe_k\rangle\right|^2\\&=\|U^*XU\|^2\\&=\text{tr}(U^*X^*UU^*XU)\\&=\text{tr}(U^*X^*XU)\\&=\text{tr}(X^*XUU^*)\\&=\text{tr}(X^*X)=\|X\|^2.
\end{align*}$$
As a different approach, we can use Parseval's identity
$$\begin{align*}
\sum_{k=1}^n\sum_{j=1}^n\left|\langle u_j,Xu_k\rangle\right|^2 &=\sum_{k=1}^n\|Xu_k\|^2\\&=\sum_{k=1}^n\sum_{j=1}^n\left|\langle e_j,Xu_k\rangle\right|^2 \\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle X^*e_j,u_k\rangle\right|^2 \\&=\sum_{j=1}^n\sum_{k=1}^n\|X^*e_j\|^2
\\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle X^*e_j,e_k\rangle\right|^2 
\\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle e_j,Xe_k\rangle\right|^2 
\\&=\sum_{j=1}^n\sum_{k=1}^n\left|X_{jk}\right|^2 =\|X\|^2.
\end{align*}$$
A: In this answer we look at a vector of $n$ coordinates as a column matrix. That is, a matrix with $n$ rows and $1$ column. First note that identity 
$$
\left\|X\right\|^2=\sum_{j=1}^n\sum_{k=1}^n|\langle u_j,Xu_k\rangle |^2\qquad (\ast)
$$
is valid when the  basis $\{u_1,\ldots,u_n\}$ of $\mathbb{C}^n$ is the canonical basis $\{e_1,\ldots,e_n\}$ of $\mathbb{C}^n$. Let any orthonormal basis $\{u_1,\ldots,u_n\}$ of $\mathbb{C}^n$. Let $U$ be the matrix whose columns are $u_1,\ldots, u_n$. That is, 
$$
U=\left[\; u_1 | \ldots |u_j|\ldots |u_n\;\right]
$$
Note that 
$$
X\cdot U= X\cdot \left[\; u_1 | \ldots |u_j|\ldots |u_n\;\right]= \left[\; X\cdot u_1 | \ldots |X\cdot u_\ell|\ldots |X\cdot u_n\;\right]
$$
and 
$$
U^\ast X\cdot U= U^\ast\left[\; X\cdot u_1 | \ldots |X\cdot u_j|\ldots |X\cdot u_n\;\right]= \big[\langle u_i, Xu_j \rangle \big]_{n\times n}
$$
Then 
$$
\left\| U^\ast X U \right\|^2
=
\left\| \big[\langle u_i, Xu_j \rangle \big]_{n\times n} \right\|^2
=
\sum_{i=1}^{n}\sum_{j=1}^{n} |\langle u_i, Xu_j \rangle| ^2
$$
If $U=(U_{ij})_{n\times n}$ and $U^\ast=(U^\ast_{ij})_{n\times n}$ then $UU^\ast=(U_{ij})_{n\times n}\cdot(U_{ij})_{n\times n}=(\sum_{k=1}^{n}U_{ik}U^\ast_{kj})_{n\times n}=I_{n\times n}$. More explicitly,
$$
\sum_{k=1}^{n}U_{ik}U^\ast_{kj}=\begin{cases} 1 & \mbox{if } i=j\\ 0 & \mbox{if } i\neq j\end{cases}
$$
Now we have 
\begin{align}
\left\|U^\ast \cdot X \cdot U\right\|^2
=&
\left\|
(U^\ast_{ij})_{n\times n}\cdot (X_{k\ell})_{n\times n}\cdot (U_{pq})_{n\times n}
\right\|^2
\\
=&
\left\|
(\sum_{\alpha=1}^nU^\ast_{i\alpha}\cdot X_{\alpha\ell})_{n\times n}\cdot (U_{pq})_{n\times n}
\right\|^2
\\
=&
\left\|
(\sum_{\beta=1}^{n}\sum_{\alpha=1}^nU^\ast_{i\alpha}\cdot X_{\alpha\beta}\cdot U_{\beta q})_{n\times n}
\right\|^2
\\
=&
\sum_{i=1}^{n}\sum_{q=1}^{n}
\sum_{\beta=1}^{n}\sum_{\alpha=1}^n
U^\ast_{i\alpha}\cdot X_{\alpha\beta}\cdot U_{\beta q}
\cdot 
\overline{U^\ast_{i\alpha}\cdot X_{\alpha\beta}\cdot U_{\beta q}}
\\
=&
\sum_{\beta=1}^{n}\sum_{\alpha=1}^n
X_{\alpha\beta}\cdot \overline{X_{\alpha\beta}}
\sum_{i=1}^{n}\sum_{q=1}^{n} 
U_{\beta q} \cdot  U^\ast_{i\alpha}\overline{U_{\beta q} \cdot  U^\ast_{i\alpha}}
\\
=&
\sum_{\beta=1}^{n}\sum_{\alpha=1}^n X_{\alpha\beta}\cdot \overline{X_{\alpha\beta}} 
\Big(
\sum_{i=q}^{n} U_{\beta q} \cdot  U^\ast_{i\alpha}\overline{U_{\beta q} \cdot  U^\ast_{i\alpha}}
+
\sum_{1\leq q<i\leq n}^{n} U_{\beta q} \cdot  U^\ast_{i\alpha}\overline{U_{\beta q} \cdot  U^\ast_{i\alpha}}
\\
&\hspace{9cm}+
\sum_{1\leq i<q\leq n}^{n} U_{\beta q} \cdot  U^\ast_{i\alpha}\overline{U_{\beta q} \cdot  U^\ast_{i\alpha}}
\Big)
\\
=&
\sum_{\beta=1}^{n}\sum_{\alpha=1}^n X_{\alpha\beta}\cdot \overline{X_{\alpha\beta}} 
\Big(
1
+
0
+
0
\Big)
\\
=&
\sum_{\beta=1}^{n}\sum_{\alpha=1}^n X_{\alpha\beta}\overline{X_{\alpha\beta}}
\\
=&
\left\|X\right\|^2
\end{align}
