{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe use the construction $V \\otimes W = L / N$ of the [[tensor product|0G31VJ5YS5]], where $v \\otimes w$ is the class of the basis vector $[v, w]$ of $L$.\n\n1. The elements of $N$ have class $0$. So $(v + v') \\otimes w = v \\otimes w + v' \\otimes w$, $v \\otimes (w + w') = v \\otimes w + v \\otimes w'$, and $(av) \\otimes w = a (v \\otimes w) = v \\otimes (aw)$.\n2. Each element of $V \\otimes W$ is the class of a finite combination $\\sum_k c_k [v_k, w_k]$. So it is $\\sum_k c_k (v_k \\otimes w_k)$, and the vectors $v \\otimes w$ [[span|0BJKVF90Z6]] $V \\otimes W$.\n3. Let $v \\in V$ and $w \\in W$. Since $(e_i)$ and $(f_j)$ are [[bases|0VTQTPQGAV]], $v = \\sum_i a_i e_i$ and $w = \\sum_j b_j f_j$ with finitely many nonzero coefficients.\n4. By step 1, $v \\otimes w = \\sum_{i,j} a_i b_j\\, (e_i \\otimes f_j)$. With step 2, the family $(e_i \\otimes f_j)$ spans $V \\otimes W$.\n5. For independence, let $\\sum_{(i,j)} c_{ij}\\, (e_i \\otimes f_j) = 0$, with only finitely many $c_{ij} \\ne 0$. Fix $(k, l) \\in I \\times J$.\n6. By [[the extension from a basis|0R8X9XA83A]], there are [[linear maps|0RJF668P1T]] $\\alpha \\colon V \\to F$ with $\\alpha(e_i) = \\delta_{ik}$ and $\\beta \\colon W \\to F$ with $\\beta(f_j) = \\delta_{jl}$. Here $\\delta$ is $1$ for equal indices and $0$ otherwise.\n7. The map $h(v, w) = \\alpha(v) \\beta(w)$ from $V \\times W$ to $F$ is bilinear, since $\\alpha$ and $\\beta$ are linear.\n8. By the [[universal property|0YDRV01R8F]], there is a linear map $\\varphi \\colon V \\otimes W \\to F$ with $\\varphi(v \\otimes w) = \\alpha(v) \\beta(w)$.\n9. In particular $\\varphi(e_i \\otimes f_j) = \\delta_{ik} \\delta_{jl}$. Applying $\\varphi$ to the relation of step 5 gives $c_{kl} = 0$.\n10. Since $(k, l)$ was arbitrary, all $c_{ij}$ vanish. So the family is [[linearly independent|0EVHJFTG9Z]].\n11. By steps 4 and 10, $(e_i \\otimes f_j)_{(i,j) \\in I \\times J}$ is a basis of $V \\otimes W$.\n\n## Idea\n\nSpanning comes from expanding $v$ and $w$ in their bases and using bilinearity. For independence, the universal property gives a linear functional that reads off one coefficient $c_{kl}$. It comes from the bilinear map that multiplies the $k$-th coordinate of $v$ by the $l$-th coordinate of $w$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "15A69"
    ],
    "name": "Proof that the tensor products of basis vectors form a basis",
    "proves": "0SH5VT8QYA",
    "sources": [
      {
        "locator": "Theorem 4.9",
        "url": "https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf"
      }
    ],
    "tag": "GMR2HXZRMR",
    "uses": [
      "0G31VJ5YS5",
      "0YDRV01R8F",
      "0VTQTPQGAV",
      "0BJKVF90Z6",
      "0EVHJFTG9Z",
      "0R8X9XA83A",
      "0RJF668P1T"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/GMR2HXZRMR/",
  "links": [
    {
      "field": "proves",
      "target": "0SH5VT8QYA"
    },
    {
      "field": "uses",
      "target": "0G31VJ5YS5"
    },
    {
      "field": "uses",
      "target": "0YDRV01R8F"
    },
    {
      "field": "uses",
      "target": "0VTQTPQGAV"
    },
    {
      "field": "uses",
      "target": "0BJKVF90Z6"
    },
    {
      "field": "uses",
      "target": "0EVHJFTG9Z"
    },
    {
      "field": "uses",
      "target": "0R8X9XA83A"
    },
    {
      "field": "uses",
      "target": "0RJF668P1T"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/GMR2HXZRMR.md",
  "msc": [
    "15A69"
  ],
  "name": "Proof that the tensor products of basis vectors form a basis",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/GMR/GMR2HXZRMR.md",
  "status": null,
  "tag": "GMR2HXZRMR",
  "uri": "m2://GMR2HXZRMR",
  "url": "https://math2.tamnd.com/wiki/proof-that-the-tensor-products-of-basis-vectors-form-a-basis/",
  "writing": {
    "tag": "GMR2HXZRMR",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}