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  "body": "\n## Proof\n\nWe compute the coefficient of $t^{n-1}$ of $p_A$ in two ways.\n\n1. Let $A$ be an $n \\times n$ matrix over $F$. The [[characteristic polynomial|039HF58KQ5]] is defined for $n \\ge 1$, so $n \\ge 1$. Assume that its characteristic polynomial is $p_A(t) = (t - \\lambda_1) \\cdots (t - \\lambda_n)$ with $\\lambda_i \\in F$.\n2. Claim: for $k \\ge 1$, $\\prod_{i=1}^{k} (t - \\lambda_i) = t^k - (\\lambda_1 + \\cdots + \\lambda_k) t^{k-1} + r_k$, where $r_k$ is $0$ or has [[degree|0DPF6WRZW6]] at most $k - 2$.\n3. For $k = 1$, the claim holds with $r_1 = 0$.\n4. Suppose the claim holds for $k$, and let $s = \\lambda_1 + \\cdots + \\lambda_k$ and $c = \\lambda_{k+1}$. Multiply both sides by $t - c$.\n5. This gives $t^{k+1} - (s + c) t^k + r_{k+1}$, with $r_{k+1} = c s\\, t^{k-1} + t\\, r_k - c\\, r_k$.\n6. Each of the three terms of $r_{k+1}$ is $0$ or has degree at most $k - 1$. So the claim holds for $k + 1$, and by induction for all $k$.\n7. By step 2 with $k = n$, the coefficient of $t^{n-1}$ in $p_A$ is $-(\\lambda_1 + \\cdots + \\lambda_n)$.\n8. By the rule for the [[trace as a coefficient|0FRYPZ6D0X]], the same coefficient is $-\\tr A$.\n9. By steps 7 and 8, $\\tr A = \\lambda_1 + \\lambda_2 + \\cdots + \\lambda_n$.\n\n## Idea\n\nMultiplying out $(t - \\lambda_1) \\cdots (t - \\lambda_n)$ gives $t^n - (\\lambda_1 + \\cdots + \\lambda_n) t^{n-1} + \\cdots$. The same coefficient of $t^{n-1}$ is $-\\tr A$ when $\\det(t I - A)$ is expanded. Comparing the two gives the result.\n",
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    "name": "Proof that the trace is the sum of the eigenvalues by comparing coefficients",
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