[proofplan] We expand $\det(tI_n-A)$ by the Leibniz formula and separate the identity permutation from the non-identity permutations. The identity permutation produces the product $\prod_{i=1}^n(t-a_{ii})$, whose top two coefficients are $1$ and $-\sum_i a_{ii}$. Every non-identity permutation moves at least two indices, so its Leibniz product contains at most $n-2$ factors involving $t$ and cannot contribute to the coefficients of $t^n$ or $t^{n-1}$. Finally, evaluating at $t=0$ gives the constant coefficient, and the same Leibniz expansion gives $\det(-A)=(-1)^n\det A$. [/proofplan]

[step:Expand the determinant as a sum over permutations] Let $S_n$ denote the symmetric group on the set $\{1,\dots,n\}$, and for $\sigma \in S_n$ let $\operatorname{sgn}(\sigma) \in \{1,-1\}$ denote its sign. For each $\sigma \in S_n$, define a polynomial $P_\sigma \in k[t]$ by \begin{align*} P_\sigma(t):=\prod_{i=1}^{n}\bigl(tI_n-A\bigr)_{i,\sigma(i)}. \end{align*} By the [Leibniz formula for the determinant](/theorems/7882), \begin{align*} \chi_A(t)=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)P_\sigma(t). \end{align*} For each $i \in \{1,\dots,n\}$, the matrix entry of $tI_n-A$ is \begin{align*} \bigl(tI_n-A\bigr)_{i,\sigma(i)}= \begin{cases} t-a_{ii}, & \sigma(i)=i, \end{align*} \begin{align*} -a_{i,\sigma(i)}, & \sigma(i)\neq i. \end{cases} \end{align*} Thus the only factors in $P_\sigma(t)$ that involve $t$ are those indexed by fixed points of $\sigma$. [/step]

[step:Isolate the only contribution to the two highest coefficients]Let $\operatorname{id} \in S_n$ denote the identity permutation. For $\sigma=\operatorname{id}$, we have $\operatorname{sgn}(\operatorname{id})=1$ and \begin{align*} P_{\operatorname{id}}(t)=\prod_{i=1}^{n}(t-a_{ii}). \end{align*} In this product, the coefficient of $t^n$ is obtained by choosing $t$ from every factor, so it is $1$. The coefficient of $t^{n-1}$ is obtained by choosing $-a_{rr}$ from exactly one factor, indexed by $r \in \{1,\dots,n\}$, and choosing $t$ from all other factors. Hence that coefficient is \begin{align*} -\sum_{r=1}^{n}a_{rr}=-\operatorname{tr} A. \end{align*} Now let $\sigma \in S_n$ with $\sigma \neq \operatorname{id}$. Let $F_\sigma \subset \{1,\dots,n\}$ denote the set of fixed points of $\sigma$: \begin{align*} F_\sigma:=\{i \in \{1,\dots,n\}: \sigma(i)=i\}. \end{align*} Since $\sigma \neq \operatorname{id}$, it cannot have exactly $n-1$ fixed points: if all but one element were fixed, the remaining element would also have to map to itself because $\sigma$ is a bijection of $\{1,\dots,n\}$. Therefore \begin{align*} |F_\sigma|\leq n-2. \end{align*} Since only fixed-point factors contribute powers of $t$, the degree of $P_\sigma(t)$ is at most $|F_\sigma|$, and hence at most $n-2$. Thus no non-identity permutation contributes to the coefficient of $t^n$ or to the coefficient of $t^{n-1}$ in $\chi_A(t)$. Combining this with the identity-permutation contribution gives \begin{align*} c_n=1 \end{align*} and \begin{align*} c_{n-1}=-\operatorname{tr} A. \end{align*}[/step]

[guided]The determinant expansion is a sum over permutations, so to find the two highest coefficients we ask which permutations can produce many powers of $t$. For a fixed permutation $\sigma \in S_n$, its Leibniz term is \begin{align*} P_\sigma(t)=\prod_{i=1}^{n}\bigl(tI_n-A\bigr)_{i,\sigma(i)}. \end{align*} The factor indexed by $i$ is $t-a_{ii}$ exactly when $\sigma(i)=i$, and it is the constant $-a_{i,\sigma(i)}$ when $\sigma(i)\neq i$. Hence the number of possible $t$-factors in $P_\sigma(t)$ is exactly the number of fixed points of $\sigma$. For the identity permutation $\operatorname{id}$, every index is fixed, so the corresponding term is \begin{align*} P_{\operatorname{id}}(t)=\prod_{i=1}^{n}(t-a_{ii}). \end{align*} The coefficient of $t^n$ comes from choosing $t$ in every factor, giving coefficient $1$. The coefficient of $t^{n-1}$ comes from choosing one constant factor and choosing $t$ from the remaining $n-1$ factors. If the constant factor is chosen from the $r$-th term, it contributes $-a_{rr}$. Summing over the possible choices of $r$ gives \begin{align*} -\sum_{r=1}^{n}a_{rr}=-\operatorname{tr} A. \end{align*} It remains to check that no other permutation can affect these two coefficients. Let $\sigma \in S_n$ be non-identity, and define \begin{align*} F_\sigma:=\{i \in \{1,\dots,n\}: \sigma(i)=i\}. \end{align*} Because $\sigma$ is a bijection, it cannot move exactly one element. Indeed, if $n-1$ elements were fixed and $r$ were the only remaining element, then the values of $\sigma$ on the fixed elements already occupy all elements except $r$, so bijectivity forces $\sigma(r)=r$, a contradiction. Therefore a non-identity permutation moves at least two elements, which is the same as saying \begin{align*} |F_\sigma|\leq n-2. \end{align*} Since $P_\sigma(t)$ has a $t$-dependent factor only at indices in $F_\sigma$, its degree is at most $|F_\sigma|$, hence at most $n-2$. Such a polynomial has zero coefficient of $t^n$ and zero coefficient of $t^{n-1}$. Therefore the identity permutation is the only contributor to the two highest coefficients, and we obtain \begin{align*} c_n=1 \end{align*} and \begin{align*} c_{n-1}=-\operatorname{tr} A. \end{align*}[/guided]

[step:Evaluate at zero to identify the constant coefficient] The constant coefficient of $\chi_A(t)$ is its value at $t=0$, so \begin{align*} c_0=\chi_A(0)=\det(0I_n-A)=\det(-A). \end{align*} Using the Leibniz formula again, \begin{align*} \det(-A)=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}(-a_{i,\sigma(i)}). \end{align*} For each $\sigma \in S_n$, the product has exactly $n$ scalar factors multiplied by $-1$, so \begin{align*} \prod_{i=1}^{n}(-a_{i,\sigma(i)})=(-1)^n\prod_{i=1}^{n}a_{i,\sigma(i)}. \end{align*} Therefore \begin{align*} \det(-A)=(-1)^n\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}a_{i,\sigma(i)}=(-1)^n\det A. \end{align*} Thus \begin{align*} c_0=(-1)^n\det A. \end{align*} [/step]

[step:Conclude the displayed form of the characteristic polynomial] We have proved that the coefficient of $t^n$ in $\chi_A(t)$ is $1$, the coefficient of $t^{n-1}$ is $-\operatorname{tr} A$, and the constant coefficient is $(-1)^n\det A$. Since $c_n=1 \neq 0$ in the field $k$, the polynomial $\chi_A(t)$ is monic of degree $n$. Hence its coefficient expansion has the form \begin{align*} \chi_A(t)=t^n-(\operatorname{tr} A)t^{n-1}+\cdots+(-1)^n\det A. \end{align*} This proves all asserted coefficient identities and the claimed form of the [characteristic polynomial](/page/Characteristic%20Polynomial). [/step]