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I have a matrix with the following structure

$$\left[\begin{array}{cccccc|c} -1 & 1-b & 0 & \dots & 0 & 0 & b \\ 0 & -1 & 1-b & \dots & 0 & 0 & b \\ \cdots \\ 0 & 0 & 0 & \dots &-1 & 1-b & b \\ 0 & 0 & 0 & \dots & 0 & -1 & 1 \\ \hline 0 & 0 & 0 & \dots & 0 & 0 & -1 \end{array}\right]$$

I have to find the inverse of this matrix. I begin by using the block matrix inversion formula

$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&-\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\\-\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&\quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\end{bmatrix}$$

where

$$\mathbf{A} = \begin{bmatrix} -1 & 1-b & 0 & \dots & 0 & 0 \\ 0 & -1 & 1-b & \dots & 0 & 0 \\ \cdots \\ 0 & 0 & 0 & \dots &-1 & 1-b \\ 0 & 0 & 0 & \dots & 0 & -1 \end{bmatrix}$$

$$\mathbf{B} = \begin{bmatrix} b \\ b \\ \vdots \\ b \\ 1 \end{bmatrix}$$

$$\mathbf{C} = \begin{bmatrix} 0 & 0 & 0 & \dots & 0 & 0 \end{bmatrix}$$

$$\mathbf{D} = \begin{bmatrix} -1 \end{bmatrix}$$

Given that $\mathbf{C}$ and $\mathbf{D}$ I can simplify the formula as

$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\mathbf{A}^{-1} & \mathbf {A}^{-1} \mathbf {B}\\ \mathbf{0} & -1\end{bmatrix}$$

The only part left is to solve for $\mathbf{A}$ which I believe can be called an upper bidiagonal Toeplitz matrix. Unfortunately, I have not been able to find a formula to compute the inverse for the same.

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You can notice that $A=-I + (1-b)N$ where $N$ is the standard nilpotent matrix, so

$$A^{-1} = - \left( I + (1-b)N + (1-b)^2 N^2 + \cdots + (1-b)^{n-1} N^{n-1} \right)$$

and the powers of $N$ are easily computed.

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  • $\begingroup$ Thanks a lot! Can you please share a link on how to find the powers of $N$? I tried to google it, but couldn't find a good result. $\endgroup$ – Sahil Puri May 26 '19 at 5:58
  • $\begingroup$ @SahilPuri Shift matrix. $\endgroup$ – Rodrigo de Azevedo Jun 28 at 7:39
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Elements $m_{ij}$ of the inverse matrix $M=A^{-1}$ are: \begin{align} m_{ij} &= \begin{cases} -(1-b)^{j-i}, &j\ge i, \\ 0, &j<i. \end{cases} \end{align}

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