Inverse of matrix with 1 in diagonal and some entries above them.

Question

Suppose matrix N has a,b,c above the main diagonal, and all other entries equal to $0$. that is,

$N=\begin{bmatrix} 0 & a & 0 & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & c \\ 0 & 0 & 0 & 0 \end{bmatrix}$

A is $eye(4)-N$, which is

$A=\begin{bmatrix} 1 & -a & 0 & 0 \\ 0 & 1 & -b & 0 \\ 0 & 0 & 1 & -c \\ 0 & 0 & 0 & 1 \end{bmatrix}$

What is ${A}^{-1}$? Well it's not difficult to get the answer by elimination.

${ A }^{ -1 }=\begin{bmatrix} 1 & a & ab & abc \\ 0 & 1 & b & bc \\ 0 & 0 & 1 & c \\ 0 & 0 & 0 & 1 \end{bmatrix}$

But, why ${A}^{-1}$ has a pattern of ${ A }^{ -1 }={ \left( I-N \right) }^{ -1 }=I+N+{ N }^{ 2 }+{ N }^{ 3 }$??

Well, it's not obvious to me. Is there any special properties of this matrix that I don't know? Thanks for reading here, I don't know how to be specific at the title.

Answer 1

Pay attention to the fact that $\;N^4=0\;$ , and from here and the fact that every matrix commutes with its own powers and with $\;I\;$ we get

$$I=I-N^4=(I-N)(I+N+N^2+N^3)=A(I+N+N^2+N^3)$$

and from here the equality between the inverse of $\;A\;$ and that polynomial in $\;N\;$ .

Answer 2

Remember the geometric series summation formula $1+x+x^2+\cdots=\frac1{1-x}$. This formula requires some convergence condition, but always holds for nilpotent $x$: the series breaks off to a finite sum so it becomes a simple algebraic identity. (In particular $1-x$ is always invertible for nilpotent $x$.) This gives your formula.