# How can I prove $tr(ST) = tr(TS)$ for two linear operators on a finite dimensional Hilbert space?

I would like to know how to prove that $$tr(ST) = tr(TS)$$ for two linear operators $S$ and $T$ on a finite-dimensional Hilbert space $\mathcal{H}$, given the following definition of the trace operator: $$tr(T)=\sum_{i=1}^n{\langle T e_i, e_i \rangle}$$

for an orthonormal basis $\{ e_i \}_{i=1}^n$ in $\mathcal{H}$. Is there a way to prove this without representing the operators as matrices?

• You may find worthwhile the related blog by T. Tao: The cyclic property of the trace is derived, and explicit choices of orthonormal bases are not made. Along the way you encounter a central relation connecting trace and determinant: $\,\det(1 +\epsilon A) = 1+\epsilon\operatorname{tr}A+\text{O}(\epsilon^2)\,$. – Hanno Oct 5 '18 at 5:19

## 2 Answers

I am not sure that this is different from writing an operator as a matrix, but you can argue as follows: for any operator $U$ you can write $$U(\cdot) = \sum_j U(e_j) \langle \cdot , e_j \rangle$$ and thus for the trace of $ST$ you can find: $$Tr(ST) = \sum_{i,j} \langle S(e_j), e_i \rangle \langle T(e_i), e_j \rangle$$ Now in the last formula you can commute the terms in the product. Hence the result.

It sort of boils down to looking at matrices but you may decompose the identity as: $$1 = \sum_j |e_j \rangle \langle e_j|$$ which inserted in the trace yields $${\rm tr} (ST) = \sum_i \langle e_i |ST e_i\rangle = \sum_i \sum_j \langle e_i |S e_j\rangle \langle e_j | T e_i \rangle$$ and you may interchange the sums and the two factors in the product.