{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $A$ and $B$ be $n \\times n$ complex [[positive semidefinite matrices|01MMVW8CKC]].\nThen their [[Hadamard product|SQT6QCEXZ8]] $A \\circ B$, the matrix with entries $(A \\circ B)_{ij} = a_{ij} b_{ij}$, is positive semidefinite.\n\n## Remarks\n\nBy the [[spectral theorem for Hermitian matrices|0DGZBQWZWN]], $B = \\sum_k v_k v_k^*$ for some vectors $v_k \\in \\C^n$.\nFor one vector $v$, $A \\circ (v v^*) = D A D^*$ with $D = \\operatorname{diag}(v_1, \\ldots, v_n)$, which is positive semidefinite.\nSo $A \\circ B = \\sum_k D_k A D_k^*$ is a sum of positive semidefinite matrices.\nExample: $A = \\begin{pmatrix} 2 \u0026 1 \\\\ 1 \u0026 2 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 \u0026 -1 \\\\ -1 \u0026 2 \\end{pmatrix}$ are positive definite, and $A \\circ B = \\begin{pmatrix} 2 \u0026 -1 \\\\ -1 \u0026 4 \\end{pmatrix}$ has trace $6$ and determinant $7$, so it is positive definite.\nIf $A$ and $B$ are both [[positive definite|0KX66CZMBB]], then so is $A \\circ B$.\nThe hypothesis on $B$ is needed: with $A = I_2$ and $B = \\operatorname{diag}(1, -1)$, $A \\circ B = B$ is not positive semidefinite.\n\n## Explanations\n\n### L1 human\n\nMultiply two matrices entry by entry, not by the usual row-times-column rule.\nIf both matrices are positive semidefinite, the result is again positive semidefinite.\nFor the usual product this would fail, since $AB$ need not even be Hermitian.\n\n### L2 human\n\nPositive semidefinite matrices are [[Gram matrices|00SPJSDT3Z]] $a_{ij} = \\ip{x_i}{x_j}$, and the entrywise product of two Gram matrices is the Gram matrix of the tensor products $x_i \\otimes y_i$; this gives a second proof.\nThe theorem shows that the entrywise power $(a_{ij}^k)$ and the entrywise exponential $(e^{a_{ij}})$ of a positive semidefinite matrix are positive semidefinite.\nThis is why sums, products and exponentials of positive definite kernels are again positive definite kernels in machine learning.\n",
  "commit": "5fc425f97f1f6b11da5892c2948ef8c5d8805665",
  "front": {
    "about": [
      "SQT6QCEXZ8",
      "01MMVW8CKC"
    ],
    "kind": "statement",
    "msc": [
      "15B48",
      "15A45"
    ],
    "name": "Schur product theorem",
    "role": "theorem",
    "same_as": {
      "wikidata": "Q17103152",
      "wikipedia": "Schur product theorem"
    },
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Schur_product_theorem"
      }
    ],
    "tag": "TM1RE1G7JK",
    "uses": [
      "SQT6QCEXZ8",
      "01MMVW8CKC",
      "0KX66CZMBB",
      "0DGZBQWZWN"
    ]
  },
  "kind": "statement",
  "link": "https://math2.tamnd.com/node/TM1RE1G7JK/",
  "links": [
    {
      "field": "about",
      "target": "SQT6QCEXZ8"
    },
    {
      "field": "about",
      "target": "01MMVW8CKC"
    },
    {
      "field": "uses",
      "target": "SQT6QCEXZ8"
    },
    {
      "field": "uses",
      "target": "01MMVW8CKC"
    },
    {
      "field": "uses",
      "target": "0KX66CZMBB"
    },
    {
      "field": "uses",
      "target": "0DGZBQWZWN"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/TM1RE1G7JK.md",
  "msc": [
    "15B48",
    "15A45"
  ],
  "name": "Schur product theorem",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/TM1/TM1RE1G7JK.md",
  "status": null,
  "tag": "TM1RE1G7JK",
  "uri": "m2://TM1RE1G7JK",
  "url": "https://math2.tamnd.com/wiki/schur-product-theorem/",
  "writing": {
    "tag": "TM1RE1G7JK",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}