二维逆问题/单调性

完全正性

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]. 

练习 (***). 使用高斯-约旦消元法证明每个平方完全正矩阵可以分解成以下简单类型的矩阵

${\begin{pmatrix}1&0&\ldots &0\\0&x&\ldots &0\\\vdots &\vdots &\ddots &\ldots \\0&0&\ldots &1\end{pmatrix}}{\mbox{ and}}{\begin{pmatrix}1&0&\ldots &0\\x&1&\ldots &0\\\vdots &\vdots &\ddots &\ldots \\0&0&\ldots &1\end{pmatrix}},$

其中 x > 0.

练习 (**). 使用上一个练习来证明，用完全正矩阵乘以向量会减少向量符号变化的次数。也就是说，完全正矩阵具有单调性。

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.

复合矩阵

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.

由柯西-比内公式得出

$C(MN)=C(M)(C(N)$

由于对角矩阵的复合矩阵也是对角矩阵，因此可以从原始矩阵的谱分解中获得复合矩阵的谱分解。也就是说，如果

$M=SDS^{-1},$

那么

$C(M)=C(S)C(D)C(S)^{-1}.$

练习 (*). 设 M 为 n 乘 n 方阵，C_k(M) 为其 k 阶复合矩阵。用 C_n 和 C_n-1 的元素表示 M 的逆矩阵 M^-1 的元素。

练习 (**). 假设 M 是一个方阵，具有如上的谱分解。证明其 n 阶复合矩阵 C(M) 的特征值为原始矩阵所有可能的 n 元组特征值的乘积。

谱特性

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

练习 (***). 将佩龙-弗罗贝尼乌斯定理应用于完全正矩阵的复合矩阵，以证明上面的陈述，参见 [GK]。

完全正矩阵的特征向量形成线性无关的切比雪夫系统。