[proofplan] Choose an ordered basis of $V$ and write the matrices of $S$ and $T$ in that basis. The commutator then has matrix $AB - BA$, so its trace is the sum of the diagonal entries of $AB - BA$. Expanding the diagonal entries as finite sums shows that the two double sums are the same after swapping the dummy indices, hence their difference is zero. [/proofplan] [step:Represent the commutator by the matrix $AB - BA$] Let $n := \dim_k V$. Choose an ordered basis $\mathcal{E} = (e_1,\dots,e_n)$ of $V$; if $n = 0$, this is the empty basis and all sums below are empty. Let $A = (a_{ij})_{1 \le i,j \le n} \in k^{n \times n}$ be the matrix of $S$ in the basis $\mathcal{E}$, and let $B = (b_{ij})_{1 \le i,j \le n} \in k^{n \times n}$ be the matrix of $T$ in the same basis. Thus, for each $j \in \{1,\dots,n\}$, \begin{align*} S(e_j) = \sum_{i=1}^{n} a_{ij} e_i. \end{align*} Also, \begin{align*} T(e_j) = \sum_{i=1}^{n} b_{ij} e_i. \end{align*} By the definition of matrix multiplication for composition of linear maps, the matrix of $S \circ T$ in the basis $\mathcal{E}$ is $AB$, and the matrix of $T \circ S$ in the basis $\mathcal{E}$ is $BA$. Since $[S,T] = S \circ T - T \circ S$, the matrix of $[S,T]$ in the basis $\mathcal{E}$ is \begin{align*} C := AB - BA. \end{align*} Therefore, by the definition of trace of a [linear map](/page/Linear%20Map) through any representing matrix, \begin{align*} \operatorname{tr}([S,T]) = \operatorname{tr}(C). \end{align*} [guided] We choose a basis because trace is computed from the diagonal of a matrix representative. Let $n := \dim_k V$, and choose an ordered basis $\mathcal{E} = (e_1,\dots,e_n)$ of $V$. If $n = 0$, then $\mathcal{E}$ is empty and every finite sum over $\{1,\dots,n\}$ is an empty sum, equal to $0$. Define $A = (a_{ij})_{1 \le i,j \le n} \in k^{n \times n}$ to be the matrix of the linear map $S: V \to V$ in the basis $\mathcal{E}$. This means that, for each basis vector $e_j$, \begin{align*} S(e_j) = \sum_{i=1}^{n} a_{ij} e_i. \end{align*} Define $B = (b_{ij})_{1 \le i,j \le n} \in k^{n \times n}$ to be the matrix of the linear map $T: V \to V$ in the same basis. Thus, for each $j \in \{1,\dots,n\}$, \begin{align*} T(e_j) = \sum_{i=1}^{n} b_{ij} e_i. \end{align*} The reason for using the same basis for both maps is that composition of linear maps is then represented by matrix multiplication. Let $v \in k^n$ be an arbitrary coordinate vector. In this basis, $S \circ T$ is represented by $AB$, because applying $T$ first gives the coordinate vector $Bv$, and applying $S$ next gives $A(Bv) = (AB)v$. Similarly, $T \circ S$ is represented by $BA$. The commutator is defined by \begin{align*} [S,T] := S \circ T - T \circ S. \end{align*} Therefore the matrix of $[S,T]$ in the basis $\mathcal{E}$ is \begin{align*} C := AB - BA. \end{align*} Finally, the trace of a linear map on a finite-dimensional [vector space](/page/Vector%20Space) is the trace of any matrix representing it in an ordered basis. Hence \begin{align*} \operatorname{tr}([S,T]) = \operatorname{tr}(C). \end{align*} [/guided] [/step] [step:Expand the diagonal entries and cancel the two finite double sums] For each $i \in \{1,\dots,n\}$, the $i$-th diagonal entry of $C = AB - BA$ is \begin{align*} C_{ii} = \sum_{j=1}^{n} a_{ij}b_{ji} - \sum_{j=1}^{n} b_{ij}a_{ji}. \end{align*} Thus \begin{align*} \operatorname{tr}(C) = \sum_{i=1}^{n}\sum_{j=1}^{n} a_{ij}b_{ji} - \sum_{i=1}^{n}\sum_{j=1}^{n} b_{ij}a_{ji}. \end{align*} In the second finite double sum, rename the dummy indices $i$ and $j$. Since multiplication in the field $k$ is commutative, this gives \begin{align*} \sum_{i=1}^{n}\sum_{j=1}^{n} b_{ij}a_{ji} = \sum_{i=1}^{n}\sum_{j=1}^{n} b_{ji}a_{ij} = \sum_{i=1}^{n}\sum_{j=1}^{n} a_{ij}b_{ji}. \end{align*} Therefore \begin{align*} \operatorname{tr}(C) = 0. \end{align*} [/step] [step:Conclude that the trace of the commutator vanishes] From the first step, $\operatorname{tr}([S,T]) = \operatorname{tr}(C)$. From the second step, $\operatorname{tr}(C) = 0$. Hence \begin{align*} \operatorname{tr}([S,T]) = 0. \end{align*} This proves the theorem. [/step]