Totally positive matrix (original) (raw)

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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.

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

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

det ( B ) > 0 {\displaystyle \det(\mathbf {B} )>0} $\det(\mathbf {B} )>0$

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

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

the spectral properties of kernels and matrices which are totally positive,
ordinary differential equations whose Green's function is totally positive, which arises in the theory of mechanical vibrations (by M. G. Krein and some colleagues in the mid-1930s),
the variation diminishing properties (started by I. J. Schoenberg in 1930),
Pólya frequency functions (by I. J. Schoenberg in the late 1940s and early 1950s).

Theorem. (Gantmacher, Krein, 1941)[3] If 0 < x 0 < ⋯ < x n {\displaystyle 0<x_{0}<\dots <x_{n}} $0<x_{0}<\dots <x_{n}$ 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&x_{1}&x_{1}^{2}&\dots &x_{1}^{n}\\1&x_{2}&x_{2}^{2}&\dots &x_{2}^{n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\1&x_{n}&x_{n}^{2}&\dots &x_{n}^{n}\end{bmatrix}}} $V=V(x_{0},x_{1},\cdots ,x_{n})={\begin{bmatrix}1&x_{0}&x_{0}^{2}&\dots &x_{0}^{n}\\1&x_{1}&x_{1}^{2}&\dots &x_{1}^{n}\\1&x_{2}&x_{2}^{2}&\dots &x_{2}^{n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\1&x_{n}&x_{n}^{2}&\dots &x_{n}^{n}\end{bmatrix}}$ is totally positive.

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

The case where 0 ≤ α 0 < ⋯ < α n {\displaystyle 0\leq \alpha _{0}<\dots <\alpha _{n}} $0\leq \alpha _{0}<\dots <\alpha _{n}$ are rational positive real numbers reduces to the previous case. Set p i / q i = α i {\displaystyle p_{i}/q_{i}=\alpha _{i}} $p_{i}/q_{i}=\alpha _{i}$ , then let x i ′ := x i 1 / q i {\displaystyle x'_{i}:=x_{i}^{1/q_{i}}} $x'_{i}:=x_{i}^{1/q_{i}}$ . 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}} $0\leq \alpha _{0}<\dots <\alpha _{n}$ 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}} $\alpha _{0}<\dots <\alpha _{n}$ are real numbers reduces to the previous case. Let α i ′ = α i − α 0 {\displaystyle \alpha _{i}'=\alpha _{i}-\alpha _{0}} $\alpha _{i}'=\alpha _{i}-\alpha _{0}$ , and define V i j ′ = x i α j ′ {\displaystyle V_{ij}'=x_{i}^{\alpha _{j}'}} $V_{ij}'=x_{i}^{\alpha _{j}'}$ . Now by the previous case, V ′ {\displaystyle V'} $V'$ is totally positive by noting that any minor of V {\displaystyle V} $V$ is the product of a diagonal matrix with positive entries, and a minor of V ′ {\displaystyle V'} $V'$ , so its determinant is also positive.

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

Compound matrix

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

Allan Pinkus (2009), Totally Positive Matrices, Cambridge University Press, ISBN 9780521194082
Fallat, Shaun M.; Johnson, Charles R., eds. (2011). Totally nonnegative matrices. Princeton series in applied mathematics. Princeton: Princeton University Press. ISBN 978-0-691-12157-4.
Spectral Properties of Totally Positive Kernels and Matrices, Allan Pinkus
Parametrizations of Canonical Bases and Totally Positive Matrices, Arkady Berenstein
Tensor Product Multiplicities, Canonical Bases And Totally Positive Varieties (2001), A. Berenstein, A. Zelevinsky