# On 2D Inverse Problems/Variation diminishing property

(Redirected from User:Daviddaved/Variation diminition property)

### Total positivityEdit

A totally positive matrix is a matrix in which the determinant of every square submatrix, is positive. Certain submatrices of kernels and matrices of Dirichlet-to-Neumann operators are totally positive. In fact the total positivity essentially characterizes the matrices and that allows one to obtain the valid data for discrete inverse problems from the continuous one, see [CIM], [CMM] and [IM].

Exercise (***). Use the Gauss-Jordan elimination algorithm to prove that every square totally positive matrix can be factorized into the matrices of the following simple types:
${\displaystyle {\begin{pmatrix}1&0&\ldots &0\\0&x&\ldots &0\\\vdots &\vdots &\ddots &\ldots \\0&0&\ldots &1\end{pmatrix}}{\mbox{ and}}{\begin{pmatrix}1&0&\ldots &0\\x&1&\ldots &0\\\vdots &\vdots &\ddots &\ldots \\0&0&\ldots &1\end{pmatrix}},}$

where x > 0.

Exercise (**). Use the previous exercise to show that multiplying a vector by a totally positive matrix decreases the number of sign changes in the vector. That is, totally positive matrices have variation diminishing property.
One can prove the total positivity property of restrictions of kernels of planar domains using the variation diminishing property or by approximation by planar networks. The rotation invariance and the total positivity of the kernels together are equivalent to the convolutions functions being positive-definite and are completely characterized by Bochner theorem.


### Compound matricesEdit

The compound matrix is an important construction for the study of totally positive matrices, see [GK]. For a given matrix M the compound matrix C of order n is the matrix which entries are equal to the determinants of the n by n square submatrices of M arranged in the lexicographical order. Therefore, a matrix M is totally positive if and only if its compound matrices of all orders have positive entries.


It follows from the Cauchy-Binet formula that:

${\displaystyle C(MN)=C(M)(C(N)}$

Since the compound matrix of a diagonal matrix is also diagonal, one can obtain the spectral decomposition of a compound matrix from the spectral decomposition of the original matrix. That is, if

${\displaystyle M=SDS^{-1},}$

then

${\displaystyle C(M)=C(S)C(D)C(S)^{-1}.}$
Exercise (*). Let M be an n by n square matrix and Ck(M) it's compound matrices of order k. Express the entries of the inverse M-1 of M in terms of the entries of Cn and Cn-1.
Exercise (**). Suppose, M is a square matrix that has a spectral decomposition as above. Prove that the eigenvalues of its compound matrix C(M) of order n are equal to products of all possible n-tuples of the eigenvalues of the original matrix.

### Spectral propertiesEdit

The spectrum of a square totally positive matrix is simple. That is, all its eigenvalues are positive and have multiplicity one.

Exercise (***). Use the Perron-Frobenius theorem applied to the compound matrices of a totally positive matrix to prove the statement above, see [GK].

The eigenvectors of a totally positive matrix form linearly independent Chebyshev system.