Let $S$ be a self-adjoint linear operator on an $n$-dimensional inner product space $V$. Then $$\sum_{i=1}^n\langle Su_i,u_i\rangle$$ is a constant for any orthonormal basis $\{u_1,\ldots,u_n\}$ of $V$. In fact, in terms of matrices, one easily sees that $$\sum_iu_i^\ast Su_i=\operatorname{tr}(\sum_iSu_iu_i^\ast)=\operatorname{tr}(SUU^\ast)=\operatorname{tr}(S).$$ However, is there any coordinate-free proof?


One nice approach is as follows.

Claim: If $S$ has rank $1$, then the sum will come out to $\operatorname{tr}(S)$ for any choice of orthonormal basis.

Proof: Note that any rank-1 operator can be written in the form $S(x) = \langle x,v \rangle w$ for some vectors $v,w$ (in particular: $v$ spans the image of $S$ and $w$ spans the image of $S^*$). With that, we note that for any orthonormal basis $u_i$, we have $$ \sum_{i=1}^n \langle Su_i,u_i \rangle = \sum_{i=1}^n \langle \langle u_i,v \rangle w,u_i \rangle = \sum_{i=1}^n \langle u_i,v \rangle \cdot \langle w,u_i \rangle. $$ We recognize the above as a dot-product. That is, we note that $$ \langle w,v \rangle = \langle \sum_{i=1}^n \langle w,u_i\rangle u_i , \sum_{j=1}^n \langle v,u_j\rangle u_j\rangle = \sum_{i,j = 1}^n \langle u_j,v \rangle \cdot \langle w,u_i \rangle \cdot \langle u_i,u_j\rangle = \sum_{i=1}^n \langle u_i,v \rangle \cdot \langle w,u_i \rangle. $$ Because the sum is equal to $\langle w,v \rangle$ (which are inherent to $S$), we see that it does not depend on which orthonormal basis is chosen. We see that this sum is the trace of $S$ by plugging in the standard basis $e_1,\dots,e_n$. $\qquad \square$

From there, it suffices no note that the trace is a linear function and that every $S$ can be written as a sum of rank-$1$ operators. It is clear that the trace (as defined by your sum) is a linear function. To see that every operator is a sum of rank-one operators, we could use the singular-value decomposition or polar decomposition.

If we don't want to use the existence of such decompositions, it suffices to note that for some orthonormal $u_1,\dots,u_n$ every operator $S$ can be written in the form $$ S(x) = \sum_{i,j = 1}^n \alpha_{ij}\langle x,u_i \rangle u_j $$ where $\alpha_{ij} = \langle Su_i,u_j\rangle$.

| cite | improve this answer | |

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.