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  "body": "\n## Proof\n\nWe divide with remainder and use the minimality of the degree.\n\n1. Let $m_T$ be the [[minimal polynomial|0V58FPM3EQ]] of $T$. It is [[monic|0BDX4MBKT4]], and it satisfies $m_T(T) = 0$. No nonzero polynomial of smaller degree annihilates $T$, since such a polynomial could be scaled to a monic one.\n2. Let $p \\in F[x]$ with $p(T) = 0$, so $p$ is an [[annihilating polynomial|052M4SX57M]] of $T$.\n3. By the [[division algorithm for polynomials|0MES3KS0ZC]], applied with the monic divisor $m_T$, there are $q, r \\in F[x]$ with $p = q m_T + r$ and $\\deg r \u003c \\deg m_T$.\n4. The map $f \\mapsto f(T)$ of [[evaluation at $T$|0BNEXSKNRV]] respects sums and products, because powers of $T$ commute with each other. So $p(T) = q(T) m_T(T) + r(T)$.\n5. Since $p(T) = 0$ and $m_T(T) = 0$, step 4 gives $r(T) = 0$.\n6. If $r \\ne 0$, then $r$ would be a nonzero annihilating polynomial with $\\deg r \u003c \\deg m_T$. This contradicts step 1.\n7. Hence $r = 0$ and $p = q m_T$. So $m_T$ divides $p$.\n\n## Idea\n\nDivide $p$ by $m_T$. The remainder still kills $T$, but it has smaller degree than $m_T$. The minimal polynomial has the least degree among nonzero annihilating polynomials, so the remainder must be zero.\n",
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    "name": "Proof that the minimal polynomial divides every annihilating polynomial",
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    "sources": [
      {
        "locator": "Theorem 4.4",
        "url": "https://kconrad.math.uconn.edu/blurbs/linmultialg/minpolyandappns.pdf"
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    "tag": "BRFPQM1MMW",
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      "052M4SX57M",
      "0BNEXSKNRV",
      "0MES3KS0ZC",
      "0BDX4MBKT4"
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  "msc": [
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  "name": "Proof that the minimal polynomial divides every annihilating polynomial",
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