# 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

• How do you define $\lVert X\rVert$? – José Carlos Santos Feb 10 at 16:46
• $||X||$=$\sum|X_{ij}|^2$ Sir this is Hilbert Smidt norm – MathLover Feb 10 at 16:48

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*}
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
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