inverse of an infinite matrix How to find inverse of an infinite lower triangular matrix all of whose diagonal entries are 1 and the entries of each column are given by coefficients of some power series rings?
 A: Just use gaussian elimination. You have an infinite matrix $(a_{ij})_{i,j \in \mathbb{N}}$ where $a_{ii} = 1$ for all $i$, and $a_{ij} = 0$ if $i < j$. We're looking for a (lower triangular) matrix $(b_{ij})$ such that $$
\begin{pmatrix}
  1 & 0 & \ldots \\
  a_{21} & 1 & 0 & \ldots \\
  a_{31} & a_{32} & 1 & \ddots  \\
  \vdots & \vdots & \ddots & \ddots
\end{pmatrix}
\cdot
\begin{pmatrix}
  1 & 0 & \ldots \\
  b_{21} & 1 & 0 & \ldots \\
  b_{31} & b_{32} & 1 & \ddots  \\
  \vdots & \vdots & \ddots & \ddots
\end{pmatrix}  
=
  \sum_{k=1}^\infty a_{ik} b_{kj} = \sum_{k=1}^i a_{ik} b_{kj} = \delta_{i,j} \text{.}
$$
That yields $$\begin{eqnarray}
  a_{21} b_{12} &=& \delta_{2,1} &=& 0 & \quad\Rightarrow b_{21} = \ldots\\
  a_{31} + a_{32} b_{21} + b_{31} &=& \delta_{3,1} &=& 0 &\quad\Rightarrow b_{31} = \ldots\\
  a_{32}  + b_{32} &=& \delta_{3,2} &=& 0 & \quad\Rightarrow b_{32} = \ldots\\
  a_{41} + a_{42} b_{21} + a_{43} b_{31} + b_{41} &=& \delta_{4,1} &=& 0 & \quad\Rightarrow b_{41} = \ldots
\end{eqnarray}$$
and so on.
