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Let $P = (p_{i,j})$ and $Q = (q_{i,j})$ be the $n \times n$ lower triangular matrices defined by

$$p_{i,j} = \begin{cases} \dbinom{i}{j-1} & \text{for}\ 1\le j\le i \le n\\ \hfill 0 \hfill & \text{elsewhere} \end{cases}$$

and

$$q_{i,j} = \begin{cases} \dbinom{i}{j} & \text{for}\ 1 \le j \le i \le n\\ \hfill 0 \hfill & \text{elsewhere} \end{cases}$$

For any square matrix $M=(m_{i,j})$, define $M^{alt} = (m_{i,j}^{alt})$ with entries $m_{i,j}^{alt} = (-1)^{i+j} m_{i,j}$.

The following appears to be true per Mathematica:

$$P(P^{-1})^{alt} = Q.$$

Any ideas how to go about proving it?

Thank you.

metamorphy
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Jose A Rodriguez
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  • $Q$ appears to the lower-triangular case of Pascal's matrix: https://en.wikipedia.org/wiki/Pascal_matrix. This type of matrix has been considered previously on this site, so you may wish to search around in the question archives. – Semiclassical Jan 25 '21 at 19:35

1 Answers1

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$P^{-1}$ has a closed form in terms of Bernoulli numbers: $$(P^{-1})_{i,j}=\begin{cases}\displaystyle\frac{1}{i}\binom{i}{j}B_{i-j},& 1\leqslant j\leqslant i\\\hfill 0,\hfill& \text{otherwise}\end{cases}$$

A direct way to show it is to use the definition $\sum_{n=0}^\infty B_n x^n/n!=x/(e^x-1)$. Namely, $$\sum_{k=j}^i\binom{i}{k-1}\frac{1}{k}\binom{k}{j}B_{k-j}=\sum_{k=j}^i\frac{i!\ B_{k-j}}{(i-k+1)!j!(k-j)!}=\frac{i!}{j!}S_{i-j}$$ where, using the "coefficient-of" notation, $$S_n=\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n+1-k)!}=[x^{n+1}]\frac{x}{e^x-1}(e^x-1)=\begin{cases}1,&n=0\\0,&n\neq 0\end{cases}$$ as expected. Similarly, $[P(P^{-1})^{alt}]_{i,j}=(i!/j!)T_{i-j}$ where, as expected again, $$T_n=\sum_{k=0}^n\frac{(-1)^k B_k}{k!(n+1-k)!}=[x^{n+1}]\frac{-x}{e^{-x}-1}(e^x-1)=\frac{1}{n!}.$$

metamorphy
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    Thank you. After seeing your proof I realized that it can be done in a similar spirit but dealing directly with binomial coefficients, by noticing that $P(P^{-1})^{alt} = Q$ is equivalent to $P = QP^{alt}$, because $(P^{-1})^{alt} = (P^{alt})^{-1}$ (since the mapping $M \rightarrow M^{alt}$ is an isomorphism of the general linear group $GL(n)$), thus avoiding the need to invert $P$. However you obtain a very nice expression for $P^{-1}$ in terms of Bernouilli numbers. – Jose A Rodriguez Jan 29 '21 at 11:07