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, 2019 at 17:39
  • 2
    $\begingroup$ 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, 2019 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, 2019 at 20:31
  • 1
    $\begingroup$ 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, 2019 at 21:50
  • 1
    $\begingroup$ I think the given matrix is a type of the Cauchy-like matrices. $\endgroup$
    – user0410
    Apr 6, 2019 at 10:49

1 Answer 1


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 $M$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.