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  "body": "\n## Proof\n\nWe expand $\\det(t I_n - A)$ by the Leibniz formula and find the terms of degree $n - 1$.\n\n1. Let $A = (a_{ij})$ be an $n \\times n$ matrix over $K$, and let $N = t I_n - A$, a matrix over the [[polynomial ring|0Q9EZS90GY]] $K[t]$.\n2. The entries of $N$ are $N_{ii} = t - a_{ii}$ and $N_{ij} = -a_{ij}$ for $i \\ne j$. Each diagonal entry has [[degree|0DPF6WRZW6]] at most $1$, and each other entry is a constant.\n3. By the definition of the [[characteristic polynomial|039HF58KQ5]] and the [[Leibniz formula|016PE4VZ8Q]], $p_A(t) = \\sum_{\\sigma \\in S_n} \\sgn(\\sigma) \\prod_{i=1}^{n} N_{i, \\sigma(i)}$.\n4. Let $\\sigma$ be a [[permutation|05RWFKVTJ7]] other than the identity. Then $\\sigma(i) = j$ for some $i \\ne j$.\n5. Then $\\sigma(j) \\ne j$, because $\\sigma(j) = j = \\sigma(i)$ would contradict the injectivity of $\\sigma$. So $\\sigma$ moves at least two points $i$ and $j$.\n6. So at most $n - 2$ factors of the product for $\\sigma$ are diagonal entries. The other factors are constants.\n7. A product of polynomials has degree at most the sum of their degrees. By steps 2 and 6, the term for $\\sigma$ has degree at most $n - 2$.\n8. So the coefficient of $t^{n-1}$ in $p_A$ equals the coefficient of $t^{n-1}$ in the identity term $\\prod_{i=1}^{n} (t - a_{ii})$.\n9. Claim: for $c_1, \\dots, c_k \\in K$ with $k \\ge 1$, $\\prod_{i=1}^{k} (t - c_i) = t^k - (c_1 + \\cdots + c_k) t^{k-1} + r_k$, where $r_k$ has degree at most $k - 2$ or is $0$.\n10. For $k = 1$, the claim holds with $r_1 = 0$.\n11. Suppose the claim holds for $k$, and let $s = c_1 + \\cdots + c_k$ and $c = c_{k+1}$. Multiply both sides by $t - c$.\n12. This gives $t^{k+1} - (s + c) t^k + r_{k+1}$, with $r_{k+1} = c s\\, t^{k-1} + t\\, r_k - c\\, r_k$.\n13. Each of the three terms of $r_{k+1}$ has degree at most $k - 1$, or is $0$. So the claim holds for $k + 1$, and by induction for all $k$.\n14. By steps 8 and 9 with $c_i = a_{ii}$, the coefficient of $t^{n-1}$ in $p_A$ is $-(a_{11} + \\cdots + a_{nn})$.\n15. By the definition of the [[trace|07PTPSZPME]], this is $-\\tr A$. With step 7, the identity term also gives the leading term $t^n$, so $p_A(t) = t^n - (\\tr A)\\, t^{n-1} + \\cdots$.\n\n## Idea\n\nA permutation other than the identity moves at least two indices. So its Leibniz term uses at most $n - 2$ diagonal entries $t - a_{ii}$, and it cannot reach degree $n - 1$. Only the product of the diagonal entries contributes to $t^{n-1}$, and there the coefficient is $-\\sum a_{ii}$.\n",
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