[proofplan] Choose a basis of the finite-dimensional [vector space](/page/Vector%20Space) $V$ and write the matrices of $S$ and $T$ in that basis. The matrix of the endomorphism $aS+bT$ has diagonal entries $a$ times the diagonal entries of the matrix of $S$ plus $b$ times the diagonal entries of the matrix of $T$. Summing the diagonal entries and using linearity of finite sums gives the desired identity. [/proofplan] [step:Represent the three endomorphisms in one basis] Let $n=\dim_k V$. Choose a basis $\mathcal{E}=(e_1,\dots,e_n)$ of $V$ if $n\geq 1$; if $n=0$, let $\mathcal{E}$ be the empty basis. Define the $k$-linear endomorphism \begin{align*} U: V \to V,\qquad x \mapsto aS(x)+bT(x). \end{align*} Let $A=(A_{ij})$, $B=(B_{ij})$, and $C=(C_{ij})$ denote the matrices of $S$, $T$, and $U$, respectively, with respect to the basis $\mathcal{E}$. Thus, for each $j\in\{1,\dots,n\}$, \begin{align*} S(e_j)=\sum_{i=1}^{n} A_{ij}e_i,\qquad T(e_j)=\sum_{i=1}^{n} B_{ij}e_i,\qquad U(e_j)=\sum_{i=1}^{n} C_{ij}e_i. \end{align*} When $n=0$, all displayed sums are empty sums. [/step] [step:Identify the matrix of $aS+bT$ as $aA+bB$] Fix $j\in\{1,\dots,n\}$. By the definition of $U$ and the $k$-linearity of scalar multiplication and addition in $V$, \begin{align*} U(e_j)=aS(e_j)+bT(e_j). \end{align*} Substituting the coordinate expansions of $S(e_j)$ and $T(e_j)$ gives \begin{align*} U(e_j)=a\sum_{i=1}^{n}A_{ij}e_i+b\sum_{i=1}^{n}B_{ij}e_i. \end{align*} Using the distributive law in the vector space $V$, \begin{align*} U(e_j)=\sum_{i=1}^{n}(aA_{ij}+bB_{ij})e_i. \end{align*} Since coordinates in a basis are unique, $C_{ij}=aA_{ij}+bB_{ij}$ for every $i,j\in\{1,\dots,n\}$. Hence $C=aA+bB$. [guided] The purpose of choosing one basis is that all three endomorphisms can be compared entry by entry. Let $j\in\{1,\dots,n\}$ be fixed. The vector $e_j$ is a basis vector, and the endomorphism $U=aS+bT$ is defined by \begin{align*} U(x)=aS(x)+bT(x) \end{align*} for every $x\in V$. Therefore \begin{align*} U(e_j)=aS(e_j)+bT(e_j). \end{align*} Now expand $S(e_j)$ and $T(e_j)$ in the basis $\mathcal{E}$. By definition of the matrix entries $A_{ij}$ and $B_{ij}$, \begin{align*} S(e_j)=\sum_{i=1}^{n}A_{ij}e_i \end{align*} and \begin{align*} T(e_j)=\sum_{i=1}^{n}B_{ij}e_i. \end{align*} Substitution gives \begin{align*} U(e_j)=a\sum_{i=1}^{n}A_{ij}e_i+b\sum_{i=1}^{n}B_{ij}e_i. \end{align*} Because scalar multiplication distributes over finite sums in the vector space $V$, this becomes \begin{align*} U(e_j)=\sum_{i=1}^{n}(aA_{ij}+bB_{ij})e_i. \end{align*} On the other hand, the entries $C_{ij}$ of the matrix of $U$ are defined by \begin{align*} U(e_j)=\sum_{i=1}^{n}C_{ij}e_i. \end{align*} A basis gives unique coordinates for each vector in $V$, so the coefficient of $e_i$ in these two expansions must agree. Hence \begin{align*} C_{ij}=aA_{ij}+bB_{ij} \end{align*} for all $i,j\in\{1,\dots,n\}$. This is exactly the statement that the matrix of $aS+bT$ in the chosen basis is $aA+bB$. [/guided] [/step] [step:Sum the diagonal entries to obtain linearity of trace] By the definition of trace for an endomorphism as the trace of any representing matrix in a basis, \begin{align*} \operatorname{tr}(U)=\sum_{i=1}^{n}C_{ii}. \end{align*} Using $C_{ii}=aA_{ii}+bB_{ii}$ for each $i$, \begin{align*} \operatorname{tr}(U)=\sum_{i=1}^{n}(aA_{ii}+bB_{ii}). \end{align*} By linearity of finite sums in the field $k$, \begin{align*} \operatorname{tr}(U)=a\sum_{i=1}^{n}A_{ii}+b\sum_{i=1}^{n}B_{ii}. \end{align*} Again by the definition of trace in the same basis, \begin{align*} \sum_{i=1}^{n}A_{ii}=\operatorname{tr}(S) \end{align*} and \begin{align*} \sum_{i=1}^{n}B_{ii}=\operatorname{tr}(T). \end{align*} Therefore \begin{align*} \operatorname{tr}(aS+bT)=a\operatorname{tr}(S)+b\operatorname{tr}(T). \end{align*} For $n=0$, each trace is the empty diagonal sum, hence all three traces are $0$, and the same identity holds. This proves the theorem. [/step]