determinant inequalities

There are a number of interesting inequalities bounding the determinant of a $n\\times n$ complex matrix $A$ , where $\\rho$ is its spectral radius:

1) $\\left|\\det(A)\\right|\\leq\\rho^{n}(A)$
2) $\\left|\\det(A)\\right|\\leq\\prod_{i=1}^{n}\\left(\\sum_{j=1}^{n}\\left|a_{ij}\\right|%\\right)=\\prod_{i=1}^{n}\\|a_{i}\\|_{1}$
3) $\\left|\\det(A)\\right|\\leq\\prod_{j=1}^{n}\\left(\\sum_{i=1}^{n}\\left|a_{ij}\\right|%\\right)=\\prod_{j=1}^{n}\\|a_{j}\\|_{1}$
4) $\\left|\\det(A)\\right|\\leq\\prod_{i=1}^{n}\\left(\\sum_{j=1}^{n}\\left|a_{ij}\\right|%^{2}\\right)^{\\frac{1}{2}}=\\prod_{i=1}^{n}\\|a_{i}\\|_{2}$
5) $\\left|\\det(A)\\right|\\leq\\prod_{j=1}^{n}\\left(\\sum_{i=1}^{n}\\left|a_{ij}\\right|%^{2}\\right)^{\\frac{1}{2}}=\\prod_{j=1}^{n}\\|a_{j}\\|_{2}$
6) if $A$ is Hermitian positive semidefinite, $\\det(A)\\leq\\prod_{i=1}^{n}a_{ii}$ , with equality if and only if $A$ is diagonal.

Inequalities 4)-6) are known as ”Hadamard’s inequalities”.

(Note that inequalities 2)-5) may suggest the idea that such inequalities could hold: $\\left|\\det(A)\\right|\\leq\\prod_{i=1}^{n}\\|a_{i}\\|_{p}$ or $\\left|\\det(A)\\right|\\leq\\prod_{j=1}^{n}\\|a_{j}\\|_{p}$ for any $p\\in\\mathbf{N}$ ; however, this is not true, as one can easily see with $A=\\begin{bmatrix}1&1\\\\-1&1\\end{bmatrix}$ and $p=3$ . Actually, inequalities 2)-5) give the best possible estimate of this kind.)

Proofs:

1) $\\left|\\det(A)\\right|=\\left|\\prod_{i=1}^{n}\\lambda_{i}\\right|=\\prod_{i=1}^{n}%\\left|\\lambda_{i}\\right|\\leq\\prod_{i=1}^{n}\\rho(A)=\\rho^{n}(A).$

2) If $A$ is singular, the thesis is trivial. Let then $\\det(A)\eq 0$ . Let’s define $B=DA$ , $D=diag(d_{11},d_{22},\\cdots,d_{nn})$ , $d_{ii}=\\left(\\sum_{j=1}^{n}\\left|a_{ij}\\right|\\right)^{-1}$ . (Note that $d_{ii}$ exist for any $i$ , because $\\det(A)\eq 0$ implies no all-zero row exists.) So $\\|B\\|_{\\infty}=\\max_{i}\\left(\\sum_{j=1}^{n}\\left|b_{ij}\\right|\\right)=1$ and, since $\\rho(B)\\leq\\|B\\|_{\\infty}$ , we have:

$\\left|\\det(B)\\right|=\\left|\\det(D)\\right|\\left|\\det(A)\\right|=\\left(\\prod_{i=1%}^{n}\\sum_{j=1}^{n}\\left|a_{ij}\\right|\\right)^{-1}\\left|\\det(A)\\right|\\leq\\rho%^{n}(B)\\leq\\|B\\|_{\\infty}^{n}=1$ ,

from which:

$\\left|\\det(A)\\right|\\leq\\prod_{i=1}^{n}\\left(\\sum_{j=1}^{n}\\left|a_{ij}\\right|%\\right).$

3) Same as 2), but applied to $A^{T}$ .

4)-6) See related proofs attached to ”Hadamard’s inequalities”.

单词	DeterminantInequalities
释义	determinant inequalities There are a number of interesting inequalities bounding the determinant of a $n\\times n$ complex matrix $A$ , where $\\rho$ is its spectral radius: 1) $\\left\|\\det(A)\\right\|\\leq\\rho^{n}(A)$ 2) $\\left\|\\det(A)\\right\|\\leq\\prod_{i=1}^{n}\\left(\\sum_{j=1}^{n}\\left\|a_{ij}\\right\|%\\right)=\\prod_{i=1}^{n}\\\|a_{i}\\\|_{1}$ 3) $\\left\|\\det(A)\\right\|\\leq\\prod_{j=1}^{n}\\left(\\sum_{i=1}^{n}\\left\|a_{ij}\\right\|%\\right)=\\prod_{j=1}^{n}\\\|a_{j}\\\|_{1}$ 4) $\\left\|\\det(A)\\right\|\\leq\\prod_{i=1}^{n}\\left(\\sum_{j=1}^{n}\\left\|a_{ij}\\right\|%^{2}\\right)^{\\frac{1}{2}}=\\prod_{i=1}^{n}\\\|a_{i}\\\|_{2}$ 5) $\\left\|\\det(A)\\right\|\\leq\\prod_{j=1}^{n}\\left(\\sum_{i=1}^{n}\\left\|a_{ij}\\right\|%^{2}\\right)^{\\frac{1}{2}}=\\prod_{j=1}^{n}\\\|a_{j}\\\|_{2}$ 6) if $A$ is Hermitian positive semidefinite, $\\det(A)\\leq\\prod_{i=1}^{n}a_{ii}$ , with equality if and only if $A$ is diagonal. Inequalities 4)-6) are known as ”Hadamard’s inequalities”. (Note that inequalities 2)-5) may suggest the idea that such inequalities could hold: $\\left\|\\det(A)\\right\|\\leq\\prod_{i=1}^{n}\\\|a_{i}\\\|_{p}$ or $\\left\|\\det(A)\\right\|\\leq\\prod_{j=1}^{n}\\\|a_{j}\\\|_{p}$ for any $p\\in\\mathbf{N}$ ; however, this is not true, as one can easily see with $A=\\begin{bmatrix}1&1\\\\-1&1\\end{bmatrix}$ and $p=3$ . Actually, inequalities 2)-5) give the best possible estimate of this kind.) Proofs: 1) $\\left\|\\det(A)\\right\|=\\left\|\\prod_{i=1}^{n}\\lambda_{i}\\right\|=\\prod_{i=1}^{n}%\\left\|\\lambda_{i}\\right\|\\leq\\prod_{i=1}^{n}\\rho(A)=\\rho^{n}(A).$ 2) If $A$ is singular, the thesis is trivial. Let then $\\det(A)\eq 0$ . Let’s define $B=DA$ , $D=diag(d_{11},d_{22},\\cdots,d_{nn})$ , $d_{ii}=\\left(\\sum_{j=1}^{n}\\left\|a_{ij}\\right\|\\right)^{-1}$ . (Note that $d_{ii}$ exist for any $i$ , because $\\det(A)\eq 0$ implies no all-zero row exists.) So $\\\|B\\\|_{\\infty}=\\max_{i}\\left(\\sum_{j=1}^{n}\\left\|b_{ij}\\right\|\\right)=1$ and, since $\\rho(B)\\leq\\\|B\\\|_{\\infty}$ , we have: $\\left\|\\det(B)\\right\|=\\left\|\\det(D)\\right\|\\left\|\\det(A)\\right\|=\\left(\\prod_{i=1%}^{n}\\sum_{j=1}^{n}\\left\|a_{ij}\\right\|\\right)^{-1}\\left\|\\det(A)\\right\|\\leq\\rho%^{n}(B)\\leq\\\|B\\\|_{\\infty}^{n}=1$ , from which: $\\left\|\\det(A)\\right\|\\leq\\prod_{i=1}^{n}\\left(\\sum_{j=1}^{n}\\left\|a_{ij}\\right\|%\\right).$ 3) Same as 2), but applied to $A^{T}$ . 4)-6) See related proofs attached to ”Hadamard’s inequalities”.
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