Inverse of a triangular matrix in a statistical problem Can any one give to me idea how to solve this problem?
Find the inverse of the triangular matrix T, where
$ T =\left[ \begin{array}{ccc}
I & J & J \\ 0 & I & J \\ 0 & 0 & I \end{array}\right] $
and each sub-matrix is of order K $\times$ K and J is a matrix with each element equal to $+1$?
This question is related to linear models 
Thank you
 A: I suppose that $I$ denotes identity matrix.
The matrix $T$ can be rewritten as $T=I-A$ where
$$A=\begin{pmatrix}0&-J&-J\\0&0&-J&\\0&0&0\end{pmatrix}.$$
If you know about Neumann series, then $S=I+A+A^2+A^3+\dots$ seems as a reasonable candidate for $T^{-1}$, assuming that this series converges.
In this case we have
$$A^2=\begin{pmatrix}0&0&J^2\\0&0&0\\0&0&0\end{pmatrix}$$ 
and $A^3=A^4=\dots=A^{3+k}=0$.
So the inverse should be
$$T^{-1}=I+A+A^2=\begin{pmatrix}I&-J&J^2-J\\0&I&-J&\\0&0&I\end{pmatrix}.$$
We can check, using multiplication of block matrices, that for these two matrices we indeed get
$$\begin{pmatrix}I&J&J\\0&I&J&\\0&0&I\end{pmatrix}\cdot\begin{pmatrix}I&-J&J^2-J\\0&I&-J&\\0&0&I\end{pmatrix}=
\begin{pmatrix}I&0&0\\0&I&0\\0&0&I\end{pmatrix}.$$
(In fact, since in this case $A^3=0$, we do not need Neumann series, it suffices if we notice that $(I-A)(I+A+A^2)=I-A^3$.)
If $J$ is a $k\times k$ matrix where each element is equal to $1$, then the result can be slightly simplified using $J^2=kJ$.
