{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nThe forward implication is a determinant computation. The converse is by induction on $n$.\n\n1. Suppose $A$ is [[similar|027KDW2K72]] to an [[upper triangular|0MKABWDRNC]] matrix $U = (u_{ij})$. [[Similar matrices have the same characteristic polynomial|0TYMASG33Y]], so $p_A = p_U$.\n2. The matrix $t I_n - U$ is upper triangular over the commutative ring $F[t]$, with diagonal entries $t - u_{ii}$. By the [[determinant of a triangular matrix|0K1KKMFHYH]], $p_U = (t - u_{11}) \\cdots (t - u_{nn})$. So $p_A$ splits over $F$.\n3. For the converse, we prove by induction on $n \\ge 1$: if $p_A$ splits over $F$, then $A$ is similar to an upper triangular matrix. For $n = 1$, every $1 \\times 1$ matrix is upper triangular.\n4. Let $n \\ge 2$, and assume the claim for size $n - 1$. The [[characteristic polynomial|039HF58KQ5]] $p_A$ is monic of degree $n$. As it splits, $p_A = \\prod_{i=1}^{s} (a_i t + b_i)$ with $a_i \\ne 0$. Comparing degrees gives $s = n$, and comparing leading coefficients gives $\\prod_i a_i = 1$. With $\\lambda_i = -b_i / a_i$, we get $p_A = (t - \\lambda_1) \\cdots (t - \\lambda_n)$.\n5. Let $L \\colon F^n \\to F^n$ be $L(x) = A x$. Its matrix in the standard basis is $A$, so $p_L = p_A$. Since $p_L(\\lambda_1) = 0$, the scalar $\\lambda_1$ is an eigenvalue of $L$ by [[the roots of the characteristic polynomial|053QVHF6N7]]. Choose an [[eigenvector|0B5H4208Z9]] $v$, so $v \\ne 0$ and $A v = \\lambda_1 v$.\n6. The set $\\set{v}$ is linearly independent, since $v \\ne 0$. By [[extension to a basis|0M4SXE5WW3]], there is a basis $B' = (v, w_2, \\dots, w_n)$ of $F^n$.\n7. Let $Q$ be the matrix with columns $v, w_2, \\dots, w_n$. It is the [[change of basis matrix|0KWXYZPJKN]] from the standard basis to $B'$. The [[matrix of $L$|0YYMY95FWS]] in the standard basis is $A$.\n8. By the [[change of basis formula|0MQXVZEBMT]], $Q$ is invertible and $Q^{-1} A Q = [L]_{B'}$. Since $L(v) = \\lambda_1 v$, the first column of $[L]_{B'}$ is $\\lambda_1 e_1$.\n9. So $Q^{-1} A Q = \\begin{pmatrix} \\lambda_1 \u0026 r \\\\ 0 \u0026 A' \\end{pmatrix}$ for some $1 \\times (n-1)$ row $r$ and some $(n-1) \\times (n-1)$ matrix $A'$.\n10. By step 9 and the [[determinant of a block triangular matrix|0J9TVTNB4K]] over $F[t]$, $p_{Q^{-1} A Q} = (t - \\lambda_1) p_{A'}$. By [[invariance under similarity|0TYMASG33Y]], this equals $p_A$.\n11. So $(t - \\lambda_1) p_{A'} = (t - \\lambda_1) \\prod_{i=2}^{n} (t - \\lambda_i)$. The ring $F[t]$ is an [[integral domain|0B5Z1EC66J]], so [[cancellation|0GJM24EXP9]] gives $p_{A'} = \\prod_{i=2}^{n} (t - \\lambda_i)$. Thus $p_{A'}$ splits.\n12. By the induction hypothesis, there is an invertible $(n-1) \\times (n-1)$ matrix $R$ such that $U' = R^{-1} A' R$ is upper triangular.\n13. Let $S$ be the [[block diagonal matrix|0M0SNKZ482]] $\\operatorname{diag}(1, R)$. It is invertible, with inverse $\\operatorname{diag}(1, R^{-1})$.\n14. Block multiplication gives $S^{-1} Q^{-1} A Q S = \\begin{pmatrix} \\lambda_1 \u0026 r R \\\\ 0 \u0026 U' \\end{pmatrix}$. This matrix is upper triangular, since $U'$ is.\n15. With the invertible matrix $P = Q S$, step 14 says $P^{-1} A P$ is upper triangular. This completes the induction.\n16. Now let $F$ be [[algebraically closed|02SJG1HC32]]. We show by induction on $d \\ge 1$ that each monic $f \\in F[t]$ of degree $d$ splits. The polynomial $f$ has a root $c \\in F$, and the [[factor theorem|0K3MWA7DMC]] gives $f = (t - c) g$.\n17. Here $g$ is monic of degree $d - 1$. If $d = 1$, then $g = 1$. If $d \\ge 2$, then $g$ splits by the induction hypothesis, so $f$ splits.\n18. So over an algebraically closed field, $p_A$ splits, and $A$ is similar to an upper triangular matrix by step 15.\n19. By the [[fundamental theorem of algebra|0JP0W07Y9A]], each nonconstant complex polynomial has a complex root. So $\\C$ is algebraically closed, and step 18 applies to $\\C$.\n\n## Idea\n\nA triangular matrix visibly has a split characteristic polynomial. Conversely, a root of $p_A$ gives an eigenvector. Putting it first in a basis makes the first column $\\lambda_1 e_1$. The remaining $(n-1) \\times (n-1)$ block still has a split characteristic polynomial, so induction triangularises it.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "15A20",
      "15A21"
    ],
    "name": "Proof of triangularisation by induction on the size",
    "proves": "0363JMQ42Y",
    "sources": [
      {
        "locator": "Triangularisability",
        "url": "https://en.wikipedia.org/wiki/Triangular_matrix"
      }
    ],
    "tag": "PQ79GJDGRQ",
    "uses": [
      "039HF58KQ5",
      "0MKABWDRNC",
      "027KDW2K72",
      "0TYMASG33Y",
      "0K1KKMFHYH",
      "053QVHF6N7",
      "0B5H4208Z9",
      "0M4SXE5WW3",
      "0KWXYZPJKN",
      "0MQXVZEBMT",
      "0YYMY95FWS",
      "0J9TVTNB4K",
      "0GJM24EXP9",
      "0B5Z1EC66J",
      "0M0SNKZ482",
      "0K3MWA7DMC",
      "02SJG1HC32",
      "0JP0W07Y9A"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/PQ79GJDGRQ/",
  "links": [
    {
      "field": "proves",
      "target": "0363JMQ42Y"
    },
    {
      "field": "uses",
      "target": "039HF58KQ5"
    },
    {
      "field": "uses",
      "target": "0MKABWDRNC"
    },
    {
      "field": "uses",
      "target": "027KDW2K72"
    },
    {
      "field": "uses",
      "target": "0TYMASG33Y"
    },
    {
      "field": "uses",
      "target": "0K1KKMFHYH"
    },
    {
      "field": "uses",
      "target": "053QVHF6N7"
    },
    {
      "field": "uses",
      "target": "0B5H4208Z9"
    },
    {
      "field": "uses",
      "target": "0M4SXE5WW3"
    },
    {
      "field": "uses",
      "target": "0KWXYZPJKN"
    },
    {
      "field": "uses",
      "target": "0MQXVZEBMT"
    },
    {
      "field": "uses",
      "target": "0YYMY95FWS"
    },
    {
      "field": "uses",
      "target": "0J9TVTNB4K"
    },
    {
      "field": "uses",
      "target": "0GJM24EXP9"
    },
    {
      "field": "uses",
      "target": "0B5Z1EC66J"
    },
    {
      "field": "uses",
      "target": "0M0SNKZ482"
    },
    {
      "field": "uses",
      "target": "0K3MWA7DMC"
    },
    {
      "field": "uses",
      "target": "02SJG1HC32"
    },
    {
      "field": "uses",
      "target": "0JP0W07Y9A"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/PQ79GJDGRQ.md",
  "msc": [
    "15A20",
    "15A21"
  ],
  "name": "Proof of triangularisation by induction on the size",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/PQ7/PQ79GJDGRQ.md",
  "status": null,
  "tag": "PQ79GJDGRQ",
  "uri": "m2://PQ79GJDGRQ",
  "url": "https://math2.tamnd.com/wiki/proof-of-triangularisation-by-induction-on-the-size/",
  "writing": {
    "tag": "PQ79GJDGRQ",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}