# Prove that $\lVert X\rVert^2 =\sum_{i,j=1}^\infty\lvert\langle u_i,Xu_j\rangle\rvert^2$.

Let $X ∈ M(n, \Bbb C)$ and $\{u_1, u_2, . . . , u_n\}$ be an orthonormal basis of $\Bbb C^n$. Prove that $\lVert X\rVert^2 =\sum_{i,j=1}^\infty \lvert\langle u_i,Xu_j\rangle\rvert^2$. Here $\lVert X\rVert^2=\sum_{i,j=1}^\infty X_{ij}^2$.

I have tried like this: $\exists P$ orthogonal (as $\{u_1, u_2, . . . , u_n\}$ and $\{e_1, e_2, . . . , e_n\}$ both are orthonormal) such that $P(e_i)=u_i$ where $e_i$ is our standard basis element in $\Bbb C^n$.

Now $\sum_{i,j=1}^\infty\lvert\langle u_i,Xu_j\rangle\rvert^2=\sum_{i,j=1}^\infty\lvert\langle Pe_i,XPe_j\rangle\rvert^2=^?\sum_{i,j=1}^\infty\lvert\langle e_i,Xe_j\rangle\rvert^2$. $\ldots (1)$

Because I know that $\lvert\langle Pe_i,Pe_j\rangle\rvert=\lvert\langle e_i,e_j\rangle\rvert$ for othogonal matrix $P$ but here we have $\lvert\langle Pe_i,XPe_j\rangle\rvert$ in equation $(1)$.

I need justification.

• I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ play nicer with TeX than <, > do :) – Patrick Stevens Feb 2 '18 at 8:35
• You have n basis vectors and your sum requires an infinite amount of vectors? – Raito Feb 2 '18 at 8:56

First use the fact that for any vector $v$, $||v||^{2}= \sum _{i=1} ^{\infty} |\langle u_i ,v \rangle|^{2}$. We now have to show that $||X||^{2}= \sum_j ||X u_j||^{2}$. Now $\sum_j ||X u_j||^{2}=\sum _{j,k} |\langle X u_j , e_k \rangle|^{2} =\sum _{j,k} |\langle u_j , Y e_k \rangle|^{2}$ where Y is the adjoint of X. Hence $\sum_j ||X u_j||^{2}=\sum ||Ye_k||^{2}$. In order to replace Y by X use the following idea: in above identities we only used the facts that $\{u_j\}$ and $\{e_j\}$ were orthonormal bases. We can interchange teh roles of these and also replace X by Y. If you do this and note that the adjoint of Y is X you will see that $\sum ||Ye_k||^{2}=\sum ||X e_k||^{2}=||X||^{2}$. I have used operator theoretic notations and I hope that is OK with you.
• Why $||v||^2=\sum_{i=1}^\infty |<u_i,v>|^2$? and why $\sum _{j,k} |\langle u_j , Y e_k \rangle|^{2}=\sum ||Ye_k||^{2}$ – user152715 Feb 2 '18 at 9:58