# Writing trace of a linear operator in terms of inner products

This fact may be obvious, but say that I have a linear operator $$T: V \to V$$, where $$V$$ is finite-dimensional, and an orthonormal basis $$\{e_1, \ldots, e_n\}$$ for $$V$$. I want to write an expression for the trace of $$T$$, which is supposed to take the form \begin{align*} \text{trace}(T) = \langle Te_1, e_1 \rangle + \ldots + \langle Te_n, e_n\}. \end{align*} I am trying to understand why this holds in general. If this were the standard basis for $$\mathbb{R}^n$$ and we were using the standard dot product, $$\langle Te_i, e_i \rangle$$ would give the $$i$$th entry of the $$i$$th column, and summing would therefore sum the diagonal of the matrix. But that doesn't generalize to any basis and any inner product.

Orthonormal expansion says that $$x = \sum_{k=1}^n \langle x,e_k\rangle e_k$$for all $$x \in V$$, provided the basis $$\mathcal{B} = (e_k)_{k=1}^n$$ is orthonormal (proof: write $$x = \sum_{i=1} x_ie_i$$ for some coefficients $$x_i$$ and hit both sides with $$\langle \cdot, e_k\rangle$$ to obtain $$x_k = \langle x,e_k\rangle$$). That said, you should recall how the trace is defined:

1. pick any basis $$\mathcal{B}$$ for $$V$$.
2. write each $$T(e_j)$$ as a combination $$\sum_{i=1}^k a_{ij} e_i$$, so that you have a matrix $$[T]_{\mathcal{B}} = (a_{ij})_{i,j=1}^n$$.
3. define $${\rm tr}(T) = {\rm tr}[T]_{\mathcal{B}}$$, where in the right we have a matrix trace.
4. show that the definition above does not depend on the choice of $$\mathcal{B}$$, by using that the matrix trace is invariant under conjugation.

Now you just want to exploit the fact that the chosen $$\mathcal{B}$$ is orthonormal. Orthonormal expansion gives $$T(e_j) = \sum_{i=1}^n \langle T(e_j),e_i\rangle e_i.$$So $$[T]_{\mathcal{B}} = (\langle T(e_j),e_i\rangle)_{i,j=1}^n$$. Then $${\rm tr}(T) = \sum_{i=1}^n \langle T(e_i),e_i\rangle.$$

If the basis is not orthonormal, one needs to write $$g_{ij} = \langle e_i,e_j\rangle$$ (and $$(g_{ij})_{i,j=1}^n$$ is no longer necessarily the identity matrix). Then $$x = \sum_{i=1}^n x_i e_i \implies \langle x,e_j\rangle = \sum_{i=1}^n x_i g_{ij} \implies x_i = \sum_{j=1}^n g^{ij}\langle x,e_j\rangle,$$where $$(g^{ij})_{i,j=1}^n$$ is the inverse matrix of $$(g_{ij})_{i,j=1}^n$$. So in general we'll have $${\rm tr}(T) = \sum_{i,j=1}^n g^{ij}\langle T(e_i),e_j\rangle.$$This is not a mistake: the trace is given in terms of the inner product by a double sum, which now accounts for the fact that the basis may be non-orthonormal.

It is a bit of work to grind through, but one can show $$\operatorname{tr} ( ABC) = \operatorname{tr} ( CAB)$$ (cyclic property). Hence for any invertible $$V$$ we have $$\operatorname{tr} (V^{-1} TV ) = \operatorname{tr} (V V^{-1} T ) = \operatorname{tr} T$$.

If $$e_k$$ is the standard basis, we have $$\operatorname{tr} T = \sum_k \langle T e_k, e_k \rangle$$. Let $$v_k$$ be an orthonormal basis and $$V = [v_1 \cdots v_n ]$$.

Then $$\operatorname{tr} T = \operatorname{tr} (V^T TV) = \sum_k \langle V^TTV e_k, e_k \rangle = \sum_k \langle TV e_k,Ve_k \rangle = \sum_k \langle T v_k, v_k \rangle$$.