Prove that the sum of the modulus squared of the matrix elements of a linear operator $\hat{A}$ is independent of the complete orthonormal basis used to represent the operator.
I believe I know how to do for the trace of matrix, but I am not sure for the modulus squared. What I have done so far is look at following $S=\sum \langle\psi|\hat{A}|\psi\rangle^2$, and if I change basis say to $\phi$, I get that $S=\sum \langle\phi|\hat{A}|\phi\rangle^2$ with a little algebra, on the inside of the square (essentially the steps for trace), and just square it at the end.
Is this ok to do?
Thanks for any help!
The sum of the modulus squared of the matrix elements is the Frobenius norm $\| \cdot \|_F$ of the matrix, which can be also calculated as the trace of $A\overline A^t=AA^*$. The latter is known to be independent of orthonormal base change of $A$, which is shown by the following calculation for $S \in U(n)$:
$$SAA^*S^*=SAS^*SA^*S^*=SAS^*(SAS^*)^*,$$
thus $$\|A\|_F=tr(AA^*)=tr(SAA^*S^*)=tr((SAS^*)(SAS^*)^*)=\|SAS^*\|_F.$$