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  "body": "\n## Proof\n\nWe combine the Cayley–Hamilton theorem with the divisibility property of the minimal polynomial.\n\n1. Let $p_A$ be the [[characteristic polynomial|039HF58KQ5]] of $A$. It is monic of degree $n$. Let $m_A$ be the [[minimal polynomial|0V58FPM3EQ]] of $A$.\n2. Let $L \\colon F^n \\to F^n$ be the endomorphism $L(x) = A x$. For $p \\in F[x]$, the [[value|0BNEXSKNRV]] $p(L)$ is the map $x \\mapsto p(A) x$, because $L^k(x) = A^k x$ for all $k$.\n3. A matrix $C$ is zero if and only if $C x = 0$ for all $x \\in F^n$. So by step 2, $p(L) = 0$ if and only if $p(A) = 0$.\n4. By step 3, $L$ and $A$ have the same annihilating polynomials, so $m_A = m_L$.\n5. By the [[Cayley–Hamilton theorem|0F765KFC2J]], $p_A(A) = 0$. By step 3, $p_A(L) = 0$.\n6. Since [[the minimal polynomial divides every annihilating polynomial|0VTAD25A3G]], $m_L$ divides $p_A$. By step 4, $m_A$ divides $p_A$.\n7. Write $p_A = m_A q$ with $q \\in F[x]$. Since $p_A \\ne 0$, also $q \\ne 0$, so $\\deg q \\ge 0$.\n8. The ring $F[x]$ is an [[integral domain|0B5Z1EC66J]], since $F$ is a field. By the [[degree of a product|0T233AZ1GE]], $n = \\deg p_A = \\deg m_A + \\deg q \\ge \\deg m_A$.\n9. So $m_A$ divides $p_A$, and $\\deg m_A \\le n$.\n\n## Idea\n\nThe Cayley–Hamilton theorem says that $p_A$ annihilates $A$. Every annihilating polynomial is a multiple of the minimal polynomial. A nonzero multiple has at least the same degree, which gives $\\deg m_A \\le n$.\n",
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        "locator": "Section 4, after Example 4.5",
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