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  "body": "\n## Proof\n\nWe use the adjugate of $t I_n - A$ and compare coefficients of powers of $t$.\n\n1. Write $p_A(t) = \\det(t I_n - A) = \\sum_{i=0}^{n} c_i t^i$ with $c_i \\in K$. By the [[definition of $p_A$|039HF58KQ5]], it is monic of degree $n$, so $c_n = 1$.\n2. Case $n = 1$. Then $A = (a)$ and $p_A(t) = t - a$. So $p_A(A) = A - a I_1 = 0$. From now on let $n \\ge 2$.\n3. A matrix with entries in $K[t]$ can be written in exactly one way as $\\sum_i t^i M_i$ with matrices $M_i$ over $K$. Here entry $(j, l)$ of $M_i$ is the coefficient of $t^i$ in entry $(j, l)$. Two such matrices are equal if and only if all their coefficient matrices are equal.\n4. Let $B = \\operatorname{adj}(t I_n - A)$ be the [[adjugate|0GMPYM0NSR]] over the commutative ring $K[t]$. Each entry of $B$ is a [[cofactor|0W2QAVJQ1W]], that is, up to sign the determinant of an $(n-1) \\times (n-1)$ submatrix of $t I_n - A$.\n5. Each entry of $t I_n - A$ has degree at most $1$ in $t$. By the [[Leibniz formula|016PE4VZ8Q]], an $(n-1) \\times (n-1)$ determinant is a sum of products of $n - 1$ entries. So each entry of $B$ has degree at most $n - 1$.\n6. By steps 3 and 5, $B = \\sum_{i=0}^{n-1} t^i B_i$ with $n \\times n$ matrices $B_i$ over $K$.\n7. By the [[product of a matrix and its adjugate|0528HKWZ1Z]] over $K[t]$, $(t I_n - A) B = p_A(t) I_n$.\n8. Expanding the left side gives $\\sum_{i=0}^{n-1} t^{i+1} B_i - \\sum_{i=0}^{n-1} t^i A B_i$. The right side is $\\sum_{i=0}^{n} t^i c_i I_n$.\n9. Comparing coefficients by step 3 gives three kinds of equations. For $t^n$: $B_{n-1} = c_n I_n$. For $1 \\le i \\le n-1$: $B_{i-1} - A B_i = c_i I_n$. For $t^0$: $-A B_0 = c_0 I_n$.\n10. Multiply the equation for $t^i$ on the left by $A^i$, for each $i = 0, 1, \\dots, n$, with $A^0 = I_n$. Then add all $n + 1$ equations.\n11. The right sides add up to $\\sum_{i=0}^{n} c_i A^i$, which is the [[value $p_A(A)$|0BNEXSKNRV]].\n12. The left sides add up to $A^n B_{n-1} + \\sum_{i=1}^{n-1} (A^i B_{i-1} - A^{i+1} B_i) - A B_0$. Regrouping, this is $\\sum_{i=1}^{n} A^i B_{i-1} - \\sum_{i=0}^{n-1} A^{i+1} B_i$.\n13. Substituting $j = i + 1$ in the second sum turns it into $\\sum_{j=1}^{n} A^j B_{j-1}$. This equals the first sum, so the left sides add up to $0$.\n14. By steps 11 and 13, $p_A(A) = 0$.\n\n## Idea\n\nThe adjugate identity $(t I_n - A) \\operatorname{adj}(t I_n - A) = p_A(t) I_n$ holds over the ring $K[t]$. Writing the adjugate as a polynomial in $t$ with matrix coefficients turns it into $n + 1$ equations between constant matrices. Multiplying them by powers of $A$ and adding makes the left side telescope to $0$, while the right side becomes $p_A(A)$.\n",
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