{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe write down an explicit basis of $\\Hom_F(V, W)$ with $\\dim V \\cdot \\dim W$ elements.\n\n1. Let $n = \\dim V$ and $m = \\dim W$. Let $(b_1, \\dots, b_n)$ be a [[basis|0KCQ9Y3QKA]] of $V$ and $(c_1, \\dots, c_m)$ a basis of $W$.\n2. Write $\\delta_{jk} = 1$ if $j = k$ and $\\delta_{jk} = 0$ if $j \\ne k$.\n3. For $1 \\le i \\le m$ and $1 \\le j \\le n$, [[the extension of values on a basis|0R8X9XA83A]] gives a linear map $E_{ij} \\colon V \\to W$ with $E_{ij}(b_k) = \\delta_{jk} c_i$ for $1 \\le k \\le n$.\n4. These $mn$ maps lie in the [[space of linear maps|0X9MBPP2Z3]] $\\Hom_F(V, W)$. Its operations are pointwise.\n5. The maps $E_{ij}$ [[span|088RYKAWK9]] $\\Hom_F(V, W)$. Let $T \\in \\Hom_F(V, W)$. For each $k$, write $T(b_k) = \\sum_{i=1}^m a_{ik} c_i$ with $a_{ik} \\in F$.\n6. Let $S = \\sum_{i=1}^m \\sum_{j=1}^n a_{ij} E_{ij}$. By the pointwise operations, $S(b_k) = \\sum_{i,j} a_{ij} \\delta_{jk} c_i = \\sum_{i=1}^m a_{ik} c_i = T(b_k)$ for each $k$.\n7. So $S$ and $T$ are linear maps that agree on a basis. By the uniqueness part of [[the same result|0R8X9XA83A]], $T = S$.\n8. The maps $E_{ij}$ are [[linearly independent|0CRMGPAH83]]. Suppose $\\sum_{i,j} a_{ij} E_{ij} = 0$, the zero map, with $a_{ij} \\in F$.\n9. Evaluating at $b_k$, as in step 6, gives $\\sum_{i=1}^m a_{ik} c_i = 0$. Since $(c_i)$ is linearly independent, $a_{ik} = 0$ for all $i$. This holds for each $k$.\n10. By steps 5 to 9, $(E_{ij})_{1 \\le i \\le m,\\, 1 \\le j \\le n}$ is a basis of $\\Hom_F(V, W)$ with $mn$ elements.\n11. So $\\Hom_F(V, W)$ is [[finite-dimensional|0ZHWKQMVX5]], and its [[dimension|03GSZ8RTZJ]] is $mn = \\dim V \\cdot \\dim W$.\n\n## Idea\n\nA linear map is fixed by the images of the basis vectors of $V$, and each image has $m$ coordinates. So a linear map is described by $mn$ free scalars, the entries of its matrix. The maps $E_{ij}$ are the maps whose matrix has a single entry $1$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "15A03",
      "15A04"
    ],
    "name": "Proof of the dimension formula for the space of linear maps",
    "proves": "0Z9N1ER0W0",
    "sources": [
      {
        "locator": "Section 2.2",
        "url": "https://dec41.user.srcf.net/notes/IB_M/linear_algebra_thm_proof.pdf"
      }
    ],
    "tag": "6CZRBQ3GP1",
    "uses": [
      "0X9MBPP2Z3",
      "0R8X9XA83A",
      "0KCQ9Y3QKA",
      "0CRMGPAH83",
      "088RYKAWK9",
      "03GSZ8RTZJ",
      "0ZHWKQMVX5"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/6CZRBQ3GP1/",
  "links": [
    {
      "field": "proves",
      "target": "0Z9N1ER0W0"
    },
    {
      "field": "uses",
      "target": "0X9MBPP2Z3"
    },
    {
      "field": "uses",
      "target": "0R8X9XA83A"
    },
    {
      "field": "uses",
      "target": "0KCQ9Y3QKA"
    },
    {
      "field": "uses",
      "target": "0CRMGPAH83"
    },
    {
      "field": "uses",
      "target": "088RYKAWK9"
    },
    {
      "field": "uses",
      "target": "03GSZ8RTZJ"
    },
    {
      "field": "uses",
      "target": "0ZHWKQMVX5"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/6CZRBQ3GP1.md",
  "msc": [
    "15A03",
    "15A04"
  ],
  "name": "Proof of the dimension formula for the space of linear maps",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/6CZ/6CZRBQ3GP1.md",
  "status": null,
  "tag": "6CZRBQ3GP1",
  "uri": "m2://6CZRBQ3GP1",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-dimension-formula-for-the-space-of-linear-maps/",
  "writing": {
    "tag": "6CZRBQ3GP1",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}