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  "body": "\n## Proof\n\nWe evaluate the characteristic polynomial at $t = 0$.\n\n1. Let $A = (a_{ij})$ be an $n \\times n$ matrix over $K$. Let $p_A(t) = \\det(t I_n - A)$ be its [[characteristic polynomial|039HF58KQ5]] in the [[polynomial ring|0Q9EZS90GY]] $K[t]$.\n2. Write $p_A(t) = c_0 + c_1 t + \\cdots + c_n t^n$ with $c_k \\in K$. Then $p_A(0) = c_0$, the constant coefficient.\n3. Let $\\operatorname{ev}_0 \\colon K[t] \\to K$ be the [[evaluation homomorphism|0125QVNCFQ]] at $0$. It maps each entry of $t I_n - A$ to the same entry of $-A$.\n4. By the [[Leibniz formula|016PE4VZ8Q]], $p_A(t)$ is a sum of products of entries of $t I_n - A$, with signs $\\pm 1$.\n5. A ring homomorphism preserves sums, products and signs. So by step 4 and the Leibniz formula for $-A$, $p_A(0) = \\operatorname{ev}_0(p_A) = \\det(-A)$.\n6. By the rule for the [[determinant of a scalar multiple|01EXSMK864]] with $c = -1$, $\\det(-A) = (-1)^n \\det A$.\n7. By steps 2, 5 and 6, the constant coefficient of $p_A$ is $p_A(0) = (-1)^n \\det A$.\n\n## Idea\n\nThe constant coefficient of a polynomial is its value at $0$. Putting $t = 0$ in $\\det(t I - A)$ gives $\\det(-A)$, and the factor $-1$ comes out of each of the $n$ rows.\n",
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