Inverse of a triangular matrix of special form

Question

How should I begin when I want to get an inverse matrix from this one?

Progress

I have tried to do it explicitly for small $n$. But now I am not able to deduce the general pattern with $n$ from it. And even if I were able to, then I would have to prove that it is valid somehow.

Answer 1

One can also derive it directly knowing the inverse of a simpler matrix. Denote the original one by $A$ and let $P$ be the permutation so that $AP$ is upper triangular ($P$ is a flipped identity). Now with $D=\mathrm{diag}(1,\ldots,n)$ we get $$\tag{1} D^{-1}AP=\begin{bmatrix} 1 & 1 & \cdots & 1 \\ & 1 & \cdots & 1 \\ & & \ddots & \vdots \\ & & & 1 \end{bmatrix}=:S. $$ The inverse of the "partial summation" matrix $S$ is the well-known "difference" matrix: $$ S^{-1}= \begin{bmatrix} 1 & -1 & & & \\ & 1 & - 1 & & \\ & & \ddots & \ddots & \\ & & & \ddots & -1 \\ & & & & 1 \end{bmatrix}. $$ So $$ A^{-1}=PS^{-1}D^{-1}. $$

Answer 2

Try writing it as a product of

$$\mathop{\rm diag}(1,2,3,\ldots n)\begin{bmatrix}1 & 1& \cdots 1\\ 1& 1&\cdots 0\\ \vdots\\ 1&0&\cdots 0\end{bmatrix}$$ The diagonal matrix is trivial to invert. The second matrix is just order-reversed discrete integration (cummulative summation) matrix, where differences between neighbouring rows give the result (the inverse is the discrete difference matrix).