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The standard Hilbert matrix $H$ is given by $$H_{ij}=\frac{1}{i+j-1},$$ and it has an inverse given for example in this MO question.

Now I have encountered a matrix $M$ of similar form, namely,

$$M_{ij}=\frac{1+(-1)^{i+j}}{i+j-1}.$$

Does anyone know the explicit formula for the inverse of M? Please provide some relevant reference if available.

A more general form for this question can be found here.

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    $\begingroup$ Note that $M$ splits into a direct sum of two square matrices: $A_{ij}:={1\over i+j-1/2}$ and $B_{ij}:={1\over i+j-3/2}$. $\endgroup$
    – Pietro Majer
    Commented Feb 22, 2017 at 20:41
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    $\begingroup$ @RodrigodeAzevedo: It is a direct sum if you properly re-order the coordinate vectors: You need to take the basis $\left(e_1, e_3, e_5, \ldots, e_2, e_4, e_6, \ldots\right)$. Or, to say it in terms of matrices, you move all the columns with odd indices to the front, and all the columns with even indices to the back; then you do the same to the rows. $\endgroup$
    – darij grinberg
    Commented Feb 22, 2017 at 21:48
  • $\begingroup$ Dings: perhaps you should write a new MO question with your extended matrix. $\endgroup$
    – T. Amdeberhan
    Commented Feb 27, 2017 at 1:15

3 Answers 3

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We present a generalization and also give an explicit solution.

If $M$ is the $n\times n$ matrix $$M=\left[\frac{1+(-1)^{i+j}}{x_i-y_j}\right]_{i,j=1}^n$$ then the inverse matrix $K:=M^{-1}$ has entries given by \begin{align} K_{a,b}=\begin{cases} 2\frac{\prod_{2j-1\neq b}x_{2j-1}-y_a}{\prod_{2j-1\neq a}y_a-y_{2j-1}}\cdot \frac{\prod_{2k-1}x_b-y_{2k-1}}{\prod_{2k-1\neq b}x_{2k-1}-x_b} \qquad \text{$a, b$ are odd} \\ \,\,\,\,\,\,\,\, 2\frac{\prod_{2j\neq b}x_{2j}-y_a}{\prod_{2j\neq a}y_a-y_{2j}}\cdot \frac{\prod_{2k}x_b-y_{2k}}{\prod_{2k\neq b}x_{2k}-x_b} \qquad \,\,\,\,\,\,\,\, \text{$a, b$ are even} \\ \qquad \qquad \qquad \,\,\, 0 \qquad \qquad \qquad \qquad \text{otherwise}. \end{cases} \end{align} Convention. For instance, when $a$ is odd, the product $\prod_{2j-1\neq b}(x_{2j-1}-y_a)$ is understood as running through all odd integers from $1$ to $n$, excluding $b$.

The solution to your problem is found by replacing $x_i=i-1$ and $y_j=-j$. Hence, in this case, \begin{align} K_{i,j}=\begin{cases}\frac{2(-1)^{a+b}n_1^2}{16^{n_1-1}(2a+2b-3)} \binom{2n_1+2a-2}{2a-2}\binom{2n_1+2b-2}{2b-2}\binom{2n_1-1}{n_1-a}\binom{2n_1-1}{n_1-b} \qquad i=2a-1,\, j=2b-1 \\ \qquad \frac{(-1)^{a+b}8ab}{16^n(2a+2b-1)}\binom{2n_2+2a}{2a}\binom{2n_2+2b}{2b}\binom{2n_2}{n_2-a}\binom{2n_2}{n_2-b} \qquad \qquad i=2a, \, j=2b \\ \qquad \qquad \qquad \qquad \qquad 0 \qquad \qquad \qquad \qquad \qquad \qquad \text{otherwise} \end{cases} \end{align} where we use designating $n_1=\lfloor\frac{n+1}2\rfloor$ and $n_2=\lfloor\frac{n}2\rfloor$.

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  • $\begingroup$ sorry for nitpicking: Shouldn't the first one be $\prod_{2j-1\neq \color{red}a}(x_{2j-1}-y_a)$ and the one below that $\prod_{2j\neq \color{red}a}(x_{2j}-y_a)$? And why does the $\neq$ restriction only apply to one of the numerators, if the whole thing is symmetric? The factor $(x_b-y_b)$ is present but $(x_a-y_a)$ is not. (Edit: I see that $(x_a-y_a)$ is absent only after my suggested correction, so maybe yours is OK, it is only written in an asymmetric way) But in any case, you want to add parenthesis. $\endgroup$
    – Wolfgang
    Commented Feb 24, 2017 at 12:48
  • $\begingroup$ I guess you mean "excluding $b$" at the end. I see now, the factor $(x_b-y_a)$ must not occur twice,that's all. $\endgroup$
    – Wolfgang
    Commented Feb 24, 2017 at 12:57
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    $\begingroup$ Or to write it symmetrically: \begin{align} K_{a,b}=\begin{cases} 2(x_b-y_a)\cdot \prod\limits_{2j-1\neq b}\dfrac{y_a-x_{2j-1}}{x_b-x_{2j-1}}\cdot \prod\limits_{2j-1\neq a}\dfrac{x_b-y_{2j-1}}{y_a-y_{2j-1}} \quad \text{$a, b$ are odd}\\ 2(x_b-y_a)\cdot\quad \prod\limits_{2j\neq b}\dfrac{y_a-x_{2j}}{x_b-x_{2j}}\quad \cdot \prod\limits_{2j\neq a}\dfrac{x_b-y_{2j}}{y_a-y_{2j}} \qquad\text{$a, b$ are even}\\ \qquad\qquad\qquad \quad0 \qquad\qquad\qquad\qquad \qquad \text{otherwise}. \end{cases} \end{align} :) $\endgroup$
    – Wolfgang
    Commented Feb 24, 2017 at 13:48
  • $\begingroup$ With simulations, I find the $K_{i,j}$ provided by T. Amdeberhan is not consistent with its numerical inverse. When $a$, $b$ are odd, the result is $K_{i,j}/4$; when $a$,$b$ are even, it is $K_{i,j}*4*16^{n-n_1}$. Is that right? $\endgroup$
    – Dings
    Commented Feb 26, 2017 at 14:52
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As observed in comments, the problem is equivalent to finding the inverse of the matrices $$H_\lambda:=\Big[{1\over i+j+\lambda}\Big]_{1\le i\le m\atop 1\le j\le m},$$ for $\lambda=-1/2$ and $\lambda=-3/2$ and order $m=\lceil n/2\rceil$, respectively $m=\lfloor n/2\rfloor$. These are also particular cases of Cauchy matrices, whose inverses admit the explicit Schetcher's formula quoted in the linked question (see the wiki article on Cauchy matrices and the various answers in the linked question). As a matter of fact, in this particular case the inversion formula takes quite a simpler form, and it turns out that the entries of $H_\lambda^{-1}$ are products of linear factors $\lambda+k$, with $2\le k\le 2m$ and multiplicity not larger than $2$. Precisely

$$H^{-1}_\lambda={1\over(m-1)!^2}\bigg[(-1)^{i+j}{m-1\choose i-1}{m-1\choose j-1}(\lambda+i+1)^{\overline m}(\lambda+j+1)^{\overline m}(\lambda+i+j)^{-1}\bigg]_{1\le i\le m\atop 1\le j\le m}$$ where $x^{\overline m}:=x(x+1)\dots(x+m-1)$ denotes a rising factorial.

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These matrices have been considered here.

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