I'm doing a research on matrix integrators and I ran into a problem in one particular case. To finish my proof the last thing remaining is to prove the nonsingularity of a specific matrix $$M_n: (m_{ij} = \frac{1}{a_i - a_j}, 1\leq i \leq n, 1\leq j \leq n,i\neq j;m_{ii} = \frac{c}{a_i - b} + \sum\limits_{k\neq i, 1\leq k \leq n}\frac{1}{a_i - a_k}),$$ where all $a_i, b$ are distinct.

To be more clear I provide $$M_2 = \begin{pmatrix} \frac{c}{a_1 -b} + \frac{1}{a_1 - a_2} && \frac{1}{a_1 - a_2}\\ \frac{1}{a_2 - a_1} && \frac{c}{a_2 -b} + \frac{1}{a_2 - a_1} \end{pmatrix}$$ $$M_3 = \begin{pmatrix} \frac{c}{a_1 -b} + \frac{1}{a_1 - a_2} + \frac{1}{a_1 - a_3} && \frac{1}{a_1 - a_2} && \frac{1}{a_1 - a_3}\\ \frac{1}{a_2 - a_1} && \frac{c}{a_2 -b} + \frac{1}{a_2 - a_1} + \frac{1}{a_2 - a_3} && \frac{1}{a_2 - a_3}\\ \frac{1}{a_3 - a_1} && \frac{1}{a_3 - a_2} && \frac{c}{a_3 - b} + \frac{1}{a_3 - a_1} + \frac{1}{a_3 - a_2} \end{pmatrix}$$

For $n \leq 7$ I calculated the $det(M_n) = \frac{c(c+1)...(c + n -1)}{\prod\limits_{1\leq i\leq n}(a_i - b)}$, but I have no idea how to prove this in general case.

I my particular case $c\in \mathbb N$, so this formula will prove the nonsingularity of $M_n$.

Any ideas and tips to prove the formula, or even to prove nonsingularity of $M_n$ in some other way - are very appreciated

  • $\begingroup$ dont know whether this helps but if you multiply the matrix by its transpose you should get a symmetrical matrix $\endgroup$ – A. P Mar 29 '19 at 17:39
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    $\begingroup$ If $D_0 := \operatorname{diag}((a_i-b)^{-1})$ and $M_0 := M-cD_0$, then$$M = M_0 + cD_0 = D_0(D_0^{-1}M_0 + cI) = (M_0D_0^{-1}+cI)D_0.$$Therefore,$$\det M = \det(D_0)\cdot\det(D_0^{-1}M_0 + cI) = \det(D_0)\cdot\det(M_0D_0^{-1}+cI).$$Now, $\det(D_0)$ is your denominator. Thus, you basically have to show that $0,1,\ldots,n-1$ are the eigenvalues of $M_0D_0^{-1}$ or $D_0^{-1}M_0$ (which have the same eigenvalues anyways). Maybe it makes sense to find the eigenvectors of that matrix. One could start with finding an element in the kernel. $\endgroup$ – amsmath Mar 29 '19 at 18:17
  • $\begingroup$ Mathematica confirms your conjecture for $n=4$ and $n=5$. But for $n=5$ it takes about 1 minute on my laptop. $\endgroup$ – amsmath Mar 29 '19 at 20:31
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    $\begingroup$ I found the eigenvector for $D_0^{-1}M_0$ with respect to $\ell = 0,\ldots,n-1$. It is $v\in\mathbb R^n$ with$$v_k = \frac{(b-a_k)^\ell}{(b-a_n)^\ell}\prod_{j=1,j\neq k}^{n-1}\frac{a_j-a_n}{a_j-a_k},\quad k=1,\ldots,n-1$$and $v_n=-1$. $\endgroup$ – amsmath Mar 29 '19 at 21:50
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    $\begingroup$ I think the given matrix is a type of the Cauchy-like matrices. $\endgroup$ – user0410 Apr 6 '19 at 10:49

The answer is a development of ideas from the comments of amsmath. The decisive step forward is derivation of the equation $(7)$ below.

Given a set of $n$ pairs of complex numbers $$\{(x_1,y_2),(x_2,y_2),\dots,(x_n,y_n)\}$$ such that $$ \forall\; i\ne j:\; x_i\ne x_j, $$ define its interpolating polynomial as $$ y(x)=\sum_{i}y_i\prod_{k\ne i}\frac{x-x_k}{x_i-x_k}.\tag1 $$

Differentiating the expression over $x$ and evaluating the result at $x_i$ one obtains: $$ y'_i\equiv y'(x_i)=\sum_{j\ne i}\frac1{x_i-x_j} \left[y_i-y_j\prod_{k\ne(i,j)}\frac{x_i-x_k}{x_j-x_k}\right].\tag2 $$ or $$ \frac{y'_i}{\prod\limits_{k\ne i}x_i-x_k} =\sum_{j\ne i}\frac1{x_i-x_j}\left[\frac{y_i}{\prod\limits_{k\ne i}x_i-x_k} +\frac{y_j}{\prod\limits_{k\ne j}x_j-x_k}\right].\tag{2a} $$ Introducing $f_i=\frac{y_i}{\prod\limits_{k\ne i}x_i-x_k}$, $f'_i=\frac{y'_i}{\prod\limits_{k\ne i}x_i-x_k}$ the equation $(\text{2a})$ can be rewritten in matrix notation as: $$ \begin{pmatrix} \sum\limits_{i\ne1}\frac{1}{x_1-x_i}& \frac1{x_1-x_2}&\cdots&\frac1{x_1-x_n}\\ \frac1{x_2-x_1}& \sum\limits_{i\ne2}\frac{1}{x_2-x_i}&\cdots&\frac1{x_2-x_n}\\ \vdots& \vdots& \ddots&\vdots\\ \frac1{x_n-x_1}&\frac1{x_n-x_2}&\cdots&\sum\limits_{i\ne n}\frac{1}{x_n-x_i}\\ \end{pmatrix} \begin{pmatrix} \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f_1\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f_2\\ \vdots\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f_n\\ \end{pmatrix}= \begin{pmatrix} \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f'_1\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f'_2\\ \vdots\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f'_n\\ \end{pmatrix}.\tag3 $$ or $$ A f=f'.\tag4 $$

Assume now a special form of the interpolating polynomial: $$ y_l(x)=(\lambda x+\beta)^l,\quad \beta,\lambda\in\mathbb C,\,\lambda\ne 0;\; l\in\mathbb Z,\,0\le l<n, $$ with corresponding $f$-vector components $$ f_{li}=\frac{(\lambda x_i+\beta)^l}{\prod\limits_{k\ne i}x_i-x_k},\quad f'_{li}=\frac{\lambda l(\lambda x_i+\beta)^{l-1}}{\prod\limits_{k\ne i}x_i-x_k}.\tag5 $$

Substituting the vectors $f_l$ and $f'_l$ into $(4)$ and multiplying both sides of the resulting equation from the left by diagonal matrix $D$ with elements $$ D_{ii}=\lambda x_i+\beta,\tag6 $$ one obtains: $$ DA f_l=\lambda l f_l.\tag7 $$

From this one concludes that $f_l$ are the eigenvectors of the matrix $DA$ with corresponding eigenvalues $\epsilon_l=\lambda l$. Observe that we have found all $n$ (distinct) eigenvalues of the matrix.

Since the determinant of the matrix $DA+Iz$ is characteristic polynomial of the matrix $-DA$, one obtains: $$\begin{align} &\det (DA+I z)=\prod_{l=0}^{n-1} z+\lambda l\tag8\\ &\implies \det (A+D^{-1}z)=\det{D^{-1}}\det (DA+Iz) =\prod_{l=0}^{n-1}\frac{z+\lambda l}{\lambda x_{l+1}+\beta}.\tag9 \end{align} $$

It remains only to observe that $A+D^{-1}z$ with $\lambda=1$, $\beta=-b$, $z=c$, $x_i=a_i$ is exactly your matrix $M$.


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