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Totally positive matrix

From Wikipedia, the free encyclopedia
Not to be confused with Positive matrix and Positive-definite matrix.

In mathematics, a totally positive matrix is a square matrix in which all the minors are positive: that is, the determinant of every square submatrix is a positive number.[1] A totally positive matrix has all entries positive, so it is also a positive matrix; and it has all principal minors positive (and positive eigenvalues). A symmetric totally positive matrix is therefore also positive-definite. A totally non-negative matrix is defined similarly, except that all the minors must be non-negative (positive or zero). Some authors use "totally positive" to include all totally non-negative matrices.

Definition

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Let A = ( A i j ) i j {\displaystyle \mathbf {A} =(A_{ij})_{ij}} {\displaystyle \mathbf {A} =(A_{ij})_{ij}} be an n ×ばつ n matrix. Consider any p { 1 , 2 , ... , n } {\displaystyle p\in \{1,2,\ldots ,n\}} {\displaystyle p\in \{1,2,\ldots ,n\}} and any p ×ばつ p submatrix of the form B = ( A i k j l ) k l {\displaystyle \mathbf {B} =(A_{i_{k}j_{\ell }})_{k\ell }} {\displaystyle \mathbf {B} =(A_{i_{k}j_{\ell }})_{k\ell }} where:

1 i 1 < ... < i p n , 1 j 1 < ... < j p n . {\displaystyle 1\leq i_{1}<\ldots <i_{p}\leq n,\qquad 1\leq j_{1}<\ldots <j_{p}\leq n.} {\displaystyle 1\leq i_{1}<\ldots <i_{p}\leq n,\qquad 1\leq j_{1}<\ldots <j_{p}\leq n.}

Then A is a totally positive matrix if:[2]

det ( B ) > 0 {\displaystyle \det(\mathbf {B} )>0} 0}"/>

for all submatrices B {\displaystyle \mathbf {B} } {\displaystyle \mathbf {B} } that can be formed this way.

History

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Topics which historically led to the development of the theory of total positivity include the study of:[2]

Examples

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Theorem. (Gantmacher, Krein, 1941)[3] If 0 < x 0 < < x n {\displaystyle 0<x_{0}<\dots <x_{n}} {\displaystyle 0<x_{0}<\dots <x_{n}} are positive real numbers, then the Vandermonde matrix V = V ( x 0 , x 1 , , x n ) = [ 1 x 0 x 0 2 ... x 0 n 1 x 1 x 1 2 ... x 1 n 1 x 2 x 2 2 ... x 2 n 1 x n x n 2 ... x n n ] {\displaystyle V=V(x_{0},x_{1},\cdots ,x_{n})={\begin{bmatrix}1&x_{0}&x_{0}^{2}&\dots &x_{0}^{n}\1円&x_{1}&x_{1}^{2}&\dots &x_{1}^{n}\1円&x_{2}&x_{2}^{2}&\dots &x_{2}^{n}\\\vdots &\vdots &\vdots &\ddots &\vdots \1円&x_{n}&x_{n}^{2}&\dots &x_{n}^{n}\end{bmatrix}}} {\displaystyle V=V(x_{0},x_{1},\cdots ,x_{n})={\begin{bmatrix}1&x_{0}&x_{0}^{2}&\dots &x_{0}^{n}\1円&x_{1}&x_{1}^{2}&\dots &x_{1}^{n}\1円&x_{2}&x_{2}^{2}&\dots &x_{2}^{n}\\\vdots &\vdots &\vdots &\ddots &\vdots \1円&x_{n}&x_{n}^{2}&\dots &x_{n}^{n}\end{bmatrix}}}is totally positive.

More generally, let α 0 < < α n {\displaystyle \alpha _{0}<\dots <\alpha _{n}} {\displaystyle \alpha _{0}<\dots <\alpha _{n}} be real numbers, and let 0 < x 0 < < x n {\displaystyle 0<x_{0}<\dots <x_{n}} {\displaystyle 0<x_{0}<\dots <x_{n}} be positive real numbers, then the generalized Vandermonde matrix V i j = x i α j {\displaystyle V_{ij}=x_{i}^{\alpha _{j}}} {\displaystyle V_{ij}=x_{i}^{\alpha _{j}}} is totally positive.

Proof (sketch). It suffices to prove the case where α 0 = 0 , ... , α n = n {\displaystyle \alpha _{0}=0,\dots ,\alpha _{n}=n} {\displaystyle \alpha _{0}=0,\dots ,\alpha _{n}=n}.

The case where 0 α 0 < < α n {\displaystyle 0\leq \alpha _{0}<\dots <\alpha _{n}} {\displaystyle 0\leq \alpha _{0}<\dots <\alpha _{n}} are rational positive real numbers reduces to the previous case. Set p i / q i = α i {\displaystyle p_{i}/q_{i}=\alpha _{i}} {\displaystyle p_{i}/q_{i}=\alpha _{i}}, then let x i := x i 1 / q i {\displaystyle x'_{i}:=x_{i}^{1/q_{i}}} {\displaystyle x'_{i}:=x_{i}^{1/q_{i}}}. This shows that the matrix is a minor of a larger Vandermonde matrix, so it is also totally positive.

The case where 0 α 0 < < α n {\displaystyle 0\leq \alpha _{0}<\dots <\alpha _{n}} {\displaystyle 0\leq \alpha _{0}<\dots <\alpha _{n}} are positive real numbers reduces to the previous case by taking the limit of rational approximations.

The case where α 0 < < α n {\displaystyle \alpha _{0}<\dots <\alpha _{n}} {\displaystyle \alpha _{0}<\dots <\alpha _{n}} are real numbers reduces to the previous case. Let α i = α i α 0 {\displaystyle \alpha _{i}'=\alpha _{i}-\alpha _{0}} {\displaystyle \alpha _{i}'=\alpha _{i}-\alpha _{0}}, and define V i j = x i α j {\displaystyle V_{ij}'=x_{i}^{\alpha _{j}'}} {\displaystyle V_{ij}'=x_{i}^{\alpha _{j}'}}. Now by the previous case, V {\displaystyle V'} {\displaystyle V'} is totally positive by noting that any minor of V {\displaystyle V} {\displaystyle V} is the product of a diagonal matrix with positive entries, and a minor of V {\displaystyle V'} {\displaystyle V'}, so its determinant is also positive.

For the case where α 0 = 0 , ... , α n = n {\displaystyle \alpha _{0}=0,\dots ,\alpha _{n}=n} {\displaystyle \alpha _{0}=0,\dots ,\alpha _{n}=n}, see (Fallat & Johnson 2011 , p. 74).

See also

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References

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  1. George M. Phillips (2003), "Total Positivity", Interpolation and Approximation by Polynomials, Springer, p. 274, ISBN 9780387002156
  2. 1 2 Spectral Properties of Totally Positive Kernels and Matrices, Allan Pinkus
  3. (Fallat & Johnson 2011 , p. 74)

Further reading

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