{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe prove the two implications separately.\n\n1. Let $m_T$ be the [[minimal polynomial|0V58FPM3EQ]] of $T$, and write $\\id = \\id_V$.\n2. First let $\\lambda$ be an [[eigenvalue|0ZND7NQ970]] of $T$. So there is $v \\in V$ with $v \\ne 0$ and $T(v) = \\lambda v$.\n3. By induction on $k$, $T^k(v) = \\lambda^k v$ for all $k \\ge 0$: the case $k = 0$ is $\\id(v) = v$, and $T^{k+1}(v) = T(\\lambda^k v) = \\lambda^{k+1} v$.\n4. Write $m_T(x) = \\sum_{k=0}^{d} c_k x^k$. By the definition of the [[value at $T$|0BNEXSKNRV]] and step 3, $m_T(T)(v) = \\sum_k c_k \\lambda^k v = m_T(\\lambda) v$.\n5. Since $m_T(T) = 0$, step 4 gives $m_T(\\lambda) v = 0$. As $v \\ne 0$, we get $m_T(\\lambda) = 0$.\n6. Conversely, let $m_T(\\lambda) = 0$. By the [[factor theorem|0K3MWA7DMC]], $m_T = (x - \\lambda) q$ for some $q \\in F[x]$.\n7. The polynomial $q$ is monic, and its [[degree|0DPF6WRZW6]] is $\\deg m_T - 1$. By the minimality of $\\deg m_T$, $q(T) \\ne 0$.\n8. So there is $w \\in V$ with $u = q(T)(w) \\ne 0$.\n9. Evaluation at $T$ respects products, so $(T - \\lambda \\id)(u) = \\big((x - \\lambda) q\\big)(T)(w) = m_T(T)(w) = 0$.\n10. Hence $T(u) = \\lambda u$ with $u \\ne 0$, and $\\lambda$ is an eigenvalue of $T$.\n11. Steps 5 and 10 show that $\\lambda$ is an eigenvalue of $T$ if and only if $m_T(\\lambda) = 0$.\n\n## Idea\n\nOn an eigenvector, a polynomial in $T$ acts as the same polynomial evaluated at $\\lambda$, so $m_T(T) = 0$ forces $m_T(\\lambda) = 0$. Conversely, if $x - \\lambda$ is a factor of $m_T$, the other factor $q$ does not kill $T$. Any nonzero vector in the image of $q(T)$ is then an eigenvector for $\\lambda$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "15A18"
    ],
    "name": "Proof that the eigenvalues are the roots of the minimal polynomial",
    "proves": "0QF4ZEFG52",
    "sources": [
      {
        "locator": "Theorem 4.7",
        "url": "https://kconrad.math.uconn.edu/blurbs/linmultialg/minpolyandappns.pdf"
      }
    ],
    "tag": "RJYK29EFS2",
    "uses": [
      "0V58FPM3EQ",
      "0ZND7NQ970",
      "0BNEXSKNRV",
      "0K3MWA7DMC",
      "0DPF6WRZW6"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/RJYK29EFS2/",
  "links": [
    {
      "field": "proves",
      "target": "0QF4ZEFG52"
    },
    {
      "field": "uses",
      "target": "0V58FPM3EQ"
    },
    {
      "field": "uses",
      "target": "0ZND7NQ970"
    },
    {
      "field": "uses",
      "target": "0BNEXSKNRV"
    },
    {
      "field": "uses",
      "target": "0K3MWA7DMC"
    },
    {
      "field": "uses",
      "target": "0DPF6WRZW6"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/RJYK29EFS2.md",
  "msc": [
    "15A18"
  ],
  "name": "Proof that the eigenvalues are the roots of the minimal polynomial",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/RJY/RJYK29EFS2.md",
  "status": null,
  "tag": "RJYK29EFS2",
  "uri": "m2://RJYK29EFS2",
  "url": "https://math2.tamnd.com/wiki/proof-that-the-eigenvalues-are-the-roots-of-the-minimal-polynomial/",
  "writing": {
    "tag": "RJYK29EFS2",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}