# Invariant sum of matrix entries squared under any orthonormal basis

Call $T$ a linear operator on $V$ where $V$ is a complex vector space of dimension $n$. Fix two orthonormal bases of $V$: $\underline{e}$ = {$e_1, ..., e_n$} and $\underline{f}$ = {$f_1, ..., f_n$}. I'm trying to prove that

$$\sum_{i=1}^{n}{\|Te_i \|^2} = \sum_{i=1}^{n}{\|Tf_i \|^2} \ \ \ (*)$$

Call $A$ the matrix of $T$ under the basis $\underline{e}$ with entries $a_{ij}$ and $B$ the matrix of $T$ under $\underline{f}$ with entries $b_{ij}$. So, $\sum_{i=1}^{n}{\|Ae_i \|^2}$ = the sum of the magnitude of the column vectors of $A$ squared. Similarly, $\sum_{i=1}^{n}{\|Bf_i \|^2}$ = the sum of the magnitude of the column vectors of $B$ squared. Therefore, $(*)$ implies that $$\sum_{i=1}^{n}{\sum_{j=1}^{n}{|a_{ij}|^2}} = \sum_{i=1}^{n}{\sum_{j=1}^{n}{|b_{ij}|^2}}$$ Thus, the sum of the entries squared of the matrix of $T$ under any orthonormal basis is the same.

If trace$(T^*T) = \sum_{i=1}^{n}{\|Td_i \|^2}$ for any orthonormal basis {$d_1, ..., d_n$}, then the proof is complete because the trace is the same regardless of the basis. But I don't know how to prove that either. I'm not quite sure how to continue.

Let $T \colon V \rightarrow V$ be a linear operator on a finite dimensional complex vector space and let $\mathcal{B} = (e_i)_{i=1}^n$ be an orthonormal basis for $V$. Writing

$$Te_i = \sum_{j=1}^n \left< Te_i, e_j \right> e_j,$$

we see that

$$[T]_{\mathcal{B}} = \begin{pmatrix} \left< Te_1, e_1 \right> & \left< Te_2, e_1 \right> & \dots & \left < Te_n, e_1 \right> \\ \left< Te_1, e_2 \right> & \left< Te_2, e_2 \right> & \dots & \left< Te_n, e_2 \right> \\ \vdots & \vdots & \vdots & \vdots \\ \left< Te_1, e_n \right> & \left< Te_2, e_n \right> & \dots & \left< Te_n, e_n \right> \end{pmatrix}$$

and this shows that

$$\operatorname{trace}(T) = \sum_{i=1}^n ([T]_{\mathcal{B}})_{ii} = \sum_{i=1}^n \left< Te_i, e_i \right>$$

for any orthonormal basis $\mathcal{B}$ of $V$. Finally,

$$\operatorname{trace}(T^{*}T) = \sum_{i=1}^n \left< T^{*}Te_i, e_i \right> = \sum_{i=1}^n \left< Te_i, Te_i \right> = \sum_{i=1}^n ||Te_i||^2$$

for any orthonormal basis $\mathcal{B}$ of $V$.