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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

Martin Sleziak
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leena adam
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1 Answers1

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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 $J^2=kJ$.

Martin Sleziak
  • 53,687
  • J is a matrix with each element equal to +1 can you explain this part what mean is different what you get in the final result thank you – leena adam Feb 27 '14 at 06:27
  • @leenaadam As I have written in my answer, if $J$ is the matrix you described, then $J^2=kJ$. So instead of $J^2-J$ you can write $(k-1)J$ (i.e., the matrix, in which each entry is equal to $(k-1)$). – Martin Sleziak Feb 27 '14 at 07:41