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Let $A$ be an $(n+1)\times(n+1)$ matrix defined by $a_{ij} = (i-1)^{j-1}$, with the convention that $0^0 = 1$.

$$A = \left[\begin{matrix} 0^0 & 0^1 & 0^2 & \ldots & 0^n\\ 1^0 & 1^1 & 1^2 & \ldots & 1^n\\ 2^0 & 2^1 & 2^2 & \ldots & 2^n\\ \vdots & \vdots & \vdots &\ddots & \vdots\\ n^0 & n^1 & n^2 & \ldots & n^n\\ \end{matrix}\right] $$

What is a closed form of $A^{-1}$ ?

I don't know how to approach this problem.
Any help will be appreciated.

Henry
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  • Perhaps using Vandermonde and the cofactor matrix approach to the inverse could be useful (perhaps overkill, I haven't tried it). – Dave Sep 19 '18 at 02:21
  • See https://math.stackexchange.com/questions/698254/proof-of-vandermonde-matrix-inverse-formula. – joriki Sep 19 '18 at 07:28
  • See also: https://math.stackexchange.com/questions/992850/inverse-of-an-infinitely-large-matrix – Nick Alger Sep 23 '19 at 18:00

1 Answers1

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Your matrix is a type of Vandermonde matrix. If we denote $x_i=i$ and recall that $x_0=0$, we can write $$ A^T = \begin{pmatrix} 1&1&\dots&1\\ 0&x_1^1&\dots&x_{n}^1\\ \vdots&\vdots&&\vdots\\ 0&x_1^n&\dots&x_{n}^n \end{pmatrix} = \begin{pmatrix} 1&u\\ 0&V \end{pmatrix}, $$ where $u=(1,\dots,1)$ is the row vector of all ones and $$ V = \begin{pmatrix} x_1^1&\dots&x_{n}^1\\ \vdots&&\vdots\\ x_1^n&\dots&x_{n}^n \end{pmatrix}. $$ As long as all the $x_i$s are different, this $V$ has an explicit inverse. See this ProofWiki entry for details. The key element is $B:=V^{-1}$ given behind that link. Using it and $$ \begin{pmatrix} 1&u\\ 0&V \end{pmatrix}^{-1} = \begin{pmatrix} 1&-uV^{-1}\\ 0&V^{-1} \end{pmatrix}, $$ we get $$ A^{-1} = \begin{pmatrix} 1&0\\ -B^{T}u^T&B^{T} \end{pmatrix}. $$

To make this more explicit, you need to take the actual formula for $B=V^{-1}$. Its elements (see the link above) are $$ b_{i j} = \begin{cases} \left({-1}\right)^{j-1} \left({\dfrac{\displaystyle \sum_{\substack{1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n-j} \mathop \le n \\ m_1, \ldots, m_{n-j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n-j}} } {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_m - x_i}\right)}}\right) & \text{when } 1 \le j < n \\ \qquad \qquad \qquad \dfrac 1 {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_i - x_m}\right)} & \text{when } j = n. \end{cases} $$ As $x_i=i$, you can simplify this much further. We can first observe that $$ \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_i - x_m}\right) = \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } (i - m) = (i-1)! (n-i)! (-1)^{n-i}, $$ as this is the product of consecutive integers skipping over zero.

To get further, we need to understand the sum $$ S_j = \sum_{\substack{1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n-j} \mathop \le n \\ m_1, \ldots, m_{n-j} \mathop \ne i} } m_1 \cdots m_{n-j} . $$ Some extreme cases are simple enough to see. Clearly $$ S_1 = 1\cdot2\cdots(i-1)\cdot(i+1)\cdots n = n!i^{-1} $$ and $$ S_{n-1} = 1+2+\cdots(i-1)+(i+1)+\cdots+n = \frac12n(n+1)-i. $$ I can't see how to get a general closed form expression. It is probably possible, but will not be very neat.

One thing that is going to be easier to handle is the vector $w=B^Tu^T$. Its components are given by $w_i=\sum_{j=1}^nb_{ji}$. But clearly a neater closed form expression for $b_{ij}$ will give $w_i$ as well, so I guess finding only $w_i$ would be unsatisfactory.

Joonas Ilmavirta
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