{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe count the ordered families of linearly independent columns. Let $F = \\F_q$ be the [[finite field|08RTDHX1SK]] with $q$ elements and let $V = F^n$, so $\\abs{V} = q^n$.\n\n1. An $n \\times n$ matrix $A$ over $F$ is the same as the ordered family $(a_1, \\dots, a_n)$ of its columns, with $a_j \\in V$. For $c \\in V$ with entries $c_j$, we have $A c = \\sum_{j} c_j a_j$.\n2. Claim: $A$ is [[invertible|0TWTS88TWN]] if and only if its columns are [[linearly independent|0EVHJFTG9Z]]. First let $A$ be invertible with inverse $A^{-1}$.\n3. If $\\sum_j c_j a_j = 0$, then $A c = 0$ by step 1, so $c = A^{-1} A c = 0$. So the columns are linearly independent.\n4. Conversely, let the columns be linearly independent. The space $V$ has [[dimension $n$|0QK7FPE6VE]], so the $n$ columns form a [[basis|096V7H13H9]] of $V$.\n5. Let $e_1, \\dots, e_n$ be the standard basis vectors. By step 4, for each $i$ there is $b_i \\in V$ with $A b_i = e_i$. The matrix $B$ with columns $b_1, \\dots, b_n$ satisfies $A B = I_n$.\n6. The columns of $B$ are linearly independent: if $B c = 0$, then $c = A B c = 0$. By steps 4 and 5 applied to $B$, there is a matrix $C$ with $B C = I_n$.\n7. Then $A = A (B C) = (A B) C = C$, so $B A = I_n$. Thus $A$ is invertible, and the claim holds.\n8. By steps 1 and 7, $\\abs{\\mathrm{GL}_n(F)}$ is the number $N_n$ of linearly independent families $(u_1, \\dots, u_n)$ in $V$. For $0 \\leq k \\leq n$, let $N_k$ be the number of linearly independent families of $k$ vectors.\n9. Let $(u_1, \\dots, u_k)$ be linearly independent with [[span|0HR573CHNM]] $W$. The map $F^k \\to W$, $c \\mapsto \\sum_i c_i u_i$, is onto by the definition of the span.\n10. It is injective: equal images give $\\sum_i (c_i - c'_i) u_i = 0$, so $c = c'$ by independence. Hence $\\abs{W} = \\abs{F^k} = q^k$.\n11. With $(u_1, \\dots, u_k)$ as in step 9, let $u \\in V$. If $u \\in W$, say $u = \\sum_i c_i u_i$, then $\\sum_i c_i u_i - u = 0$ is a nontrivial relation. So $(u_1, \\dots, u_k, u)$ is dependent.\n12. If $u \\notin W$ and $\\sum_i c_i u_i + c u = 0$, then $c = 0$, since otherwise $u = -c^{-1} \\sum_i c_i u_i \\in W$. Then all $c_i = 0$ by independence. So $(u_1, \\dots, u_k, u)$ is independent exactly when $u \\notin W$.\n13. If $(u_1, \\dots, u_{k+1})$ is linearly independent, so is $(u_1, \\dots, u_k)$, since a relation among the first $k$ vectors extends by the coefficient $0$.\n14. By steps 10, 12 and 13, each independent family of $k$ vectors extends in exactly $q^n - q^k$ ways. All independent families of $k + 1$ vectors arise in this way. So $N_{k+1} = N_k (q^n - q^k)$ for $0 \\leq k \u003c n$.\n15. The empty family is linearly independent, so $N_0 = 1$. By induction on $k$ with step 14, $N_k = \\prod_{j=0}^{k-1} (q^n - q^j)$.\n16. For $k = n$ and step 8, the [[order|0SEX9F9J5Z]] of $\\mathrm{GL}_n(\\F_q)$ is $\\prod_{k=0}^{n-1} (q^n - q^k) = (q^n - 1)(q^n - q) \\cdots (q^n - q^{n-1})$.\n\n## Idea\n\nA matrix is invertible exactly when its columns are linearly independent, so we choose the columns one at a time. The $(k+1)$-th column must avoid the span of the first $k$ columns, which has $q^k$ elements. That leaves $q^n - q^k$ choices at each stage.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "20G40"
    ],
    "name": "Proof of the order of the general linear group over a finite field by counting columns",
    "proves": "0JQ1KNQW2K",
    "sources": [
      {
        "locator": "Over finite fields",
        "url": "https://en.wikipedia.org/wiki/General_linear_group"
      }
    ],
    "tag": "E61BVD6G2D",
    "uses": [
      "08RTDHX1SK",
      "0TWTS88TWN",
      "0EVHJFTG9Z",
      "0HR573CHNM",
      "0QK7FPE6VE",
      "096V7H13H9",
      "0SEX9F9J5Z"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/E61BVD6G2D/",
  "links": [
    {
      "field": "proves",
      "target": "0JQ1KNQW2K"
    },
    {
      "field": "uses",
      "target": "08RTDHX1SK"
    },
    {
      "field": "uses",
      "target": "0TWTS88TWN"
    },
    {
      "field": "uses",
      "target": "0EVHJFTG9Z"
    },
    {
      "field": "uses",
      "target": "0HR573CHNM"
    },
    {
      "field": "uses",
      "target": "0QK7FPE6VE"
    },
    {
      "field": "uses",
      "target": "096V7H13H9"
    },
    {
      "field": "uses",
      "target": "0SEX9F9J5Z"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/E61BVD6G2D.md",
  "msc": [
    "20G40"
  ],
  "name": "Proof of the order of the general linear group over a finite field by counting columns",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/E61/E61BVD6G2D.md",
  "status": null,
  "tag": "E61BVD6G2D",
  "uri": "m2://E61BVD6G2D",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-order-of-the-general-linear-group-over-a-finite-field-by-counting-columns/",
  "writing": {
    "tag": "E61BVD6G2D",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}