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So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $n\times n$ matrix $X$ and any orthonormal basis $\{u_1,u_2,...,u_n\}$ of $\mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as $$\| X \|=\bigg(\sum_{j,k=1}^n|X_{jk}|^2\bigg)^{1/2}$$ can be expressed as $$\| X \|^2=\sum_{j,k=1}^n|\langle u_j,Xu_k\rangle|^2$$

I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have $$\langle u_j,Xu_k\rangle \rightarrow\langle e_j,Xe_k\rangle$$ where it follows that $$\langle e_j,Xe_k\rangle=X_{jk}$$ However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.

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To complete you may use two things:

  1. To figure out that $$ \|X\|^2=\bigg(\sum_{j,k=1}^n|X_{jk}|^2\bigg)=\operatorname{trace}X^*X $$ ($X^*$ is the Hermitian conjugate of $X$)
  2. To know the cyclic trace property $$ \operatorname{trace}(AB)=\operatorname{trace}(BA). $$ Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1\ldots,u_n$) we have $UU^*=I$, so \begin{align} \|X\|^2&=\operatorname{trace}X^*X=\operatorname{trace}X^*XUU^*=\\ &=\operatorname{trace}U^*X^*XU=\operatorname{trace}U^*X^*UU^*XU=\|U^*XU\|^2. \end{align} Here any element of $U^*XU$ is $$ u_j^*Xu_k=\langle u_j,Xu_k\rangle. $$
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