{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nFor $1 \\le k \\le n$, let $A_k$ be the upper left $k \\times k$ submatrix of $A$, so that $\\Delta_k = \\det A_k$ is a leading [[minor|0SD528P0EK]].\n\n1. Each $A_k$ is [[Hermitian|0EKZCX4K3P]]. Indeed its $(i, j)$ entry is $a_{ij} = \\overline{a_{ji}}$, because $A$ is Hermitian.\n2. First assume that $A$ is [[positive definite|0QQCG3JWJ4]], and fix $k$. Let $z \\in \\C^k$ with $z \\ne 0$, and let $x \\in \\C^n$ be $z$ followed by $n - k$ zeros.\n3. Then $x \\ne 0$, and $x^* A x = z^* A_k z$, where $^*$ is the [[conjugate transpose|0EX045E2BP]]. So $z^* A_k z \u003e 0$, and $A_k$ is positive definite.\n4. By the [[fundamental theorem of algebra|0JP0W07Y9A]], the characteristic polynomial of $A_k$ splits over $\\C$ as $(t - \\lambda_1) \\cdots (t - \\lambda_k)$.\n5. Each $\\lambda_i$ is a root of the characteristic polynomial, so it is an [[eigenvalue|053QVHF6N7]] of $A_k$. By [[the positivity of eigenvalues|0HYC2RHD3M]] and step 3, each $\\lambda_i$ is a positive real.\n6. Since the [[determinant is the product of the eigenvalues|0C2F40MEBC]], $\\Delta_k = \\lambda_1 \\cdots \\lambda_k \u003e 0$. This proves one direction.\n7. For the converse we argue by induction on $n$. Assume $\\Delta_k \u003e 0$ for all $k$. For $n = 1$, $A = (a)$ with $a = \\Delta_1 \u003e 0$, and $x^* A x = a \\abs{x}^2 \u003e 0$ for $x \\ne 0$.\n8. Let $n \\ge 2$, and let $A' = A_{n-1}$. The leading minors of $A'$ are $\\Delta_1, \\dots, \\Delta_{n-1}$, which are positive. By step 1 and the induction hypothesis, $A'$ is positive definite.\n9. Since $A$ is Hermitian, it has the block form $A = \\begin{pmatrix} A' \u0026 b \\\\ b^* \u0026 c \\end{pmatrix}$ with $b \\in \\C^{n-1}$ and $c \\in \\R$.\n10. Since $\\det A' = \\Delta_{n-1} \\ne 0$, the matrix $A'$ is [[invertible|FBSAPTP1PR]]. Conjugating $A' A'^{-1} = I$ gives $(A'^{-1})^* A' = I$, using $(XY)^* = Y^* X^*$ from the [[transpose of a product|09GK5PS824]].\n11. By step 10, $(A'^{-1})^* = A'^{-1}$, since a matrix has only one inverse.\n12. Let $u = A'^{-1} b$ and $s = c - b^* u$. Let $P = \\begin{pmatrix} I \u0026 -u \\\\ 0 \u0026 1 \\end{pmatrix}$, where $I$ is the $(n-1) \\times (n-1)$ identity.\n13. Block multiplication gives $AP = \\begin{pmatrix} A' \u0026 b - A'u \\\\ b^* \u0026 c - b^* u \\end{pmatrix} = \\begin{pmatrix} A' \u0026 0 \\\\ b^* \u0026 s \\end{pmatrix}$.\n14. We have $P^* = \\begin{pmatrix} I \u0026 0 \\\\ -u^* \u0026 1 \\end{pmatrix}$, and $u^* = b^* (A'^{-1})^* = b^* A'^{-1}$ by step 11. So $u^* A' = b^*$.\n15. By steps 13 and 14, $P^* A P = \\begin{pmatrix} A' \u0026 0 \\\\ b^* - u^* A' \u0026 s \\end{pmatrix} = \\begin{pmatrix} A' \u0026 0 \\\\ 0 \u0026 s \\end{pmatrix}$.\n16. The matrices $P$ and $P^*$ are triangular with all diagonal entries $1$. By the [[determinant of a triangular matrix|0K1KKMFHYH]], $\\det P = \\det P^* = 1$.\n17. By the [[multiplicativity of the determinant|04CPJ6AF00]] and the [[determinant of a block triangular matrix|0J9TVTNB4K]], $\\Delta_n = \\det(P^* A P) = \\Delta_{n-1}\\, s$.\n18. So $s = \\Delta_n / \\Delta_{n-1} \u003e 0$.\n19. Let $x \\in \\C^n$ with $x \\ne 0$. By step 16, $P$ is invertible. Let $y = P^{-1} x$, so $y \\ne 0$. Write $y = (y', y_n)$ with $y' \\in \\C^{n-1}$.\n20. By step 15, $x^* A x = y^* P^* A P y = y'^* A' y' + s \\abs{y_n}^2$.\n21. Both terms are at least $0$, by steps 8 and 18. If $y' \\ne 0$, the first term is positive. Otherwise $y_n \\ne 0$, and the second term is positive.\n22. So $x^* A x \u003e 0$ for all $x \\ne 0$. As $A$ is Hermitian, $A$ is positive definite. This completes the induction and the proof.\n\n## Idea\n\nPositive definiteness passes to the top left blocks, and a positive definite matrix has positive determinant. For the converse, a change of variables splits off the last coordinate. It writes the form as the form of $A_{n-1}$ plus $s \\abs{y_n}^2$, and the determinants show that $s = \\Delta_n / \\Delta_{n-1}$ is positive.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "15B57",
      "15A15"
    ],
    "name": "Proof of Sylvester's criterion by induction and completing the square",
    "proves": "0WKEF78PMV",
    "sources": [
      {
        "locator": "Proof for the case of positive definite matrices",
        "url": "https://en.wikipedia.org/wiki/Sylvester's_criterion"
      }
    ],
    "tag": "5BZ123RQNH",
    "uses": [
      "0QQCG3JWJ4",
      "0EKZCX4K3P",
      "0EX045E2BP",
      "0SD528P0EK",
      "0JP0W07Y9A",
      "053QVHF6N7",
      "0HYC2RHD3M",
      "0C2F40MEBC",
      "FBSAPTP1PR",
      "0K1KKMFHYH",
      "0J9TVTNB4K",
      "04CPJ6AF00",
      "09GK5PS824"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/5BZ123RQNH/",
  "links": [
    {
      "field": "proves",
      "target": "0WKEF78PMV"
    },
    {
      "field": "uses",
      "target": "0QQCG3JWJ4"
    },
    {
      "field": "uses",
      "target": "0EKZCX4K3P"
    },
    {
      "field": "uses",
      "target": "0EX045E2BP"
    },
    {
      "field": "uses",
      "target": "0SD528P0EK"
    },
    {
      "field": "uses",
      "target": "0JP0W07Y9A"
    },
    {
      "field": "uses",
      "target": "053QVHF6N7"
    },
    {
      "field": "uses",
      "target": "0HYC2RHD3M"
    },
    {
      "field": "uses",
      "target": "0C2F40MEBC"
    },
    {
      "field": "uses",
      "target": "FBSAPTP1PR"
    },
    {
      "field": "uses",
      "target": "0K1KKMFHYH"
    },
    {
      "field": "uses",
      "target": "0J9TVTNB4K"
    },
    {
      "field": "uses",
      "target": "04CPJ6AF00"
    },
    {
      "field": "uses",
      "target": "09GK5PS824"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/5BZ123RQNH.md",
  "msc": [
    "15B57",
    "15A15"
  ],
  "name": "Proof of Sylvester's criterion by induction and completing the square",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/5BZ/5BZ123RQNH.md",
  "status": null,
  "tag": "5BZ123RQNH",
  "uri": "m2://5BZ123RQNH",
  "url": "https://math2.tamnd.com/wiki/proof-of-sylvesters-criterion-by-induction-and-completing-the-square/",
  "writing": {
    "tag": "5BZ123RQNH",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}