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Consider the $(n+1)\times(n+1)$ matrix $P_n=(p_{ij})_{0\le i,j\le n}$ whose $(i,j)$-th element is given by $$ p_{ij}=\binom{n-i}{j-i}\boldsymbol{1}(i\le j)=\binom{n-i}{n-j}\boldsymbol{1}(i\le j) $$ for $0\le i,j\le n$, $n\ge0$; here $\boldsymbol{1}$ denotes the indicator function.

In other words, $P_n$ is an upper-triangular matrix whose upper half (above the main diagonal) is a Pascal's triangle flipped upside down, so the triangle's top is in the matrix's lower right corner.

I would like to find the inverse of $P_n$ for any $n\ge0$. Would appreciate any tips.

The question is very similar to this one.

Tian Vlašić
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Jason
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  • $P_n$ is the Pascal's triangle matrix rotated $180^\circ$, so its inverse is the inverse of the Pascal's matrix rotated $180^\circ$, i.e. $P_n$ with alternating signs. – Alexander Burstein Jun 26 '23 at 07:44

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Let $Q=(q_{ij})_{0 \leq i,j \leq n}$ defined by $$q_{ij} = (-1)^{i+j}p_{ij}$$

For every $0 \leq i,j \leq n$, one has \begin{align*} \sum_{k=0}^n p_{ik} q_{kj} &= \sum_{k=i}^j p_{ik}q_{kj} \\ & =\sum_{k=i}^j (-1)^{k+j}\binom{n-i}{n-k}\binom{n-k}{n-j}\\ & =\sum_{k=i}^j (-1)^{k+j}\binom{n-i}{n-j}\binom{j-i}{j-k} \\ & =\binom{n-i}{n-j}(-1)^{j+i}\sum_{k=0}^{j-i} (-1)^{k}\binom{j-i}{j-i-k} \\ & =\binom{n-i}{n-j}(-1)^{j+i}(1-1)^{j-i} = \delta_{ij} \\ \end{align*}

so $Q$ is the inverse of $P$.

TheSilverDoe
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