[proofplan] Choose a basis of $V$ and write $A=(a_{ij})$ for the matrix representing $T$ in that basis. In this basis, the characteristic polynomial becomes $\det(tI_n-A)$. Expanding this determinant by the Leibniz formula, every non-identity permutation contributes degree at most $n-2$, while the identity permutation contributes $\prod_{i=1}^n(t-a_{ii})$. The coefficient of $t^{n-1}$ is therefore $-\sum_{i=1}^n a_{ii}$, which is $-\operatorname{tr}(T)$ by the definition of trace for a [linear map](/page/Linear%20Map). [/proofplan] [step:Represent the linear map by a matrix in a chosen basis] Choose a basis $\mathcal{B}=(v_1,\dots,v_n)$ of the $k$-[vector space](/page/Vector%20Space) $V$. Let $A=(a_{ij}) \in k^{n \times n}$ denote the matrix of $T$ in the basis $\mathcal{B}$, meaning that for each $j \in \{1,\dots,n\}$, \begin{align*}T(v_j)=\sum_{i=1}^n a_{ij}v_i\end{align*} In the same basis, the linear map $t\operatorname{id}_V-T$ over the [polynomial ring](/page/Polynomial%20Ring) $k[t]$ is represented by the matrix $tI_n-A$, where $I_n$ is the $n \times n$ identity matrix. Hence \begin{align*}\chi_T(t)=\det(tI_n-A)\end{align*} [/step] [step:Use the permutation expansion to isolate possible degree $n-1$ terms] Let $S_n$ denote the symmetric group on $\{1,\dots,n\}$, let $\operatorname{sgn}(\sigma)$ denote the sign of a permutation $\sigma \in S_n$, let $\operatorname{id}$ denote the identity permutation of $\{1,\dots,n\}$, and let $\delta_{ij} \in k$ denote the Kronecker delta, so $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ otherwise. By the Leibniz formula for the determinant applied in the polynomial ring $k[t]$, \begin{align*}\det(tI_n-A)=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n(t\delta_{i,\sigma(i)}-a_{i,\sigma(i)})\end{align*} For a permutation $\sigma \in S_n$, the factor indexed by $i$ contains a term involving $t$ exactly when $\sigma(i)=i$. Thus the degree in $t$ of the product \begin{align*}\prod_{i=1}^n(t\delta_{i,\sigma(i)}-a_{i,\sigma(i)})\end{align*} is at most the number of fixed points of $\sigma$. If $\sigma \neq \operatorname{id}$, then $\sigma$ has at most $n-2$ fixed points: if it fixed $n-1$ elements, the remaining element would also have to be fixed because $\sigma$ is a bijection. Therefore every non-identity permutation contributes a polynomial of degree at most $n-2$. [guided] We now identify which terms in the determinant expansion can affect the coefficient of $t^{n-1}$. Let $S_n$ denote the symmetric group on $\{1,\dots,n\}$, let $\operatorname{sgn}(\sigma)$ denote the sign of $\sigma \in S_n$, let $\operatorname{id}$ denote the identity permutation of $\{1,\dots,n\}$, and let $\delta_{ij}$ be the Kronecker delta, so $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ otherwise. The Leibniz formula in the polynomial ring $k[t]$ gives \begin{align*}\det(tI_n-A)=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n(t\delta_{i,\sigma(i)}-a_{i,\sigma(i)})\end{align*} For a fixed permutation $\sigma$, inspect the factor $t\delta_{i,\sigma(i)}-a_{i,\sigma(i)}$. If $\sigma(i)=i$, then $\delta_{i,\sigma(i)}=1$, so the factor is $t-a_{ii}$ and can contribute one power of $t$. If $\sigma(i)\neq i$, then $\delta_{i,\sigma(i)}=0$, so the factor is $-a_{i,\sigma(i)}$ and contributes no power of $t$. Hence the largest possible power of $t$ in the product for $\sigma$ is the number of indices fixed by $\sigma$. The key point is that a non-identity permutation cannot fix exactly $n-1$ elements. Indeed, if $\sigma$ fixed all elements except possibly one element $r$, then bijectivity forces $\sigma(r)=r$ as well, because all other values are already occupied by the fixed elements. Therefore a non-identity permutation has at most $n-2$ fixed points. Consequently every non-identity permutation contributes only terms of degree at most $n-2$, so no non-identity permutation can contribute to the coefficient of $t^{n-1}$. [/guided] [/step] [step:Compute the degree $n-1$ coefficient from the identity permutation] The only contribution to the coefficient of $t^{n-1}$ comes from the identity permutation. For $\sigma=\operatorname{id}_{\{1,\dots,n\}}$, the corresponding term in the Leibniz expansion is \begin{align*} \prod_{i=1}^n(t-a_{ii}). \end{align*} In this product, the coefficient of $t^{n-1}$ is obtained by choosing $-a_{jj}$ from exactly one factor and $t$ from all remaining factors. Summing over the possible choices of $j \in \{1,\dots,n\}$ gives \begin{align*} [t^{n-1}]\prod_{i=1}^n(t-a_{ii})=-\sum_{j=1}^n a_{jj}. \end{align*} Here $[t^{n-1}]p(t)$ denotes the coefficient of $t^{n-1}$ in a polynomial $p(t) \in k[t]$. Therefore \begin{align*} c_{n-1}=-\sum_{j=1}^n a_{jj}. \end{align*} This computation also covers the case $n=1$, where the product has one factor and the coefficient of $t^0$ is $-a_{11}$. [/step] [step:Identify the diagonal sum with the trace of $T$] By definition, the trace of the matrix $A=(a_{ij})$ is \begin{align*} \operatorname{tr}(A)=\sum_{j=1}^n a_{jj}. \end{align*} Since $A$ represents $T$ in the basis $\mathcal{B}$ and the trace of a linear map is the trace of any representing matrix, we have \begin{align*} \operatorname{tr}(T)=\operatorname{tr}(A)=\sum_{j=1}^n a_{jj}. \end{align*} Combining this with the coefficient computation gives \begin{align*} c_{n-1}=-\operatorname{tr}(T). \end{align*} This proves the theorem. [/step]