# Finding the norm of a matrix given an orthonormal basis

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.

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 $$u_j^*Xu_k=\langle u_j,Xu_k\rangle.$$