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 $V$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.