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Consider the matrix $$A = \begin{bmatrix} n_{+} & n_1 & n_2 & n_3 & \cdots & n_{r-1} \\ n_1 & n_1 & 0 & 0 & \cdots & 0 \\ n_2 & 0 & n_2 & 0 & \cdots & 0 \\ n_3 & 0 & 0 & n_3 & \cdots & 0 \\ \vdots & \vdots & \vdots &\vdots & \ddots & \vdots \\ n_{r-1} & 0 & 0 & 0 & \cdots & n_{r-1} \\ \end{bmatrix} \in M_{r \times r}(\mathbb{R})\text{.}$$ where $n_{+} = \sum_{i=1}^{r}n_i$ (all $n_i > 0$ and $n_r$ is just another constant greater than $0$). Is there a formula for the inverse of this matrix?

I can see it's a symmetric matrix, but beyond that, I'm not sure what else I could do to find the inverse.

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  • $\begingroup$ The block matrix inversion formula should be helpful here. $\endgroup$ – JimmyK4542 Dec 16 '15 at 20:19
  • $\begingroup$ @JimmyK4542 How would you suggest blocking $A$? $\endgroup$ – Clarinetist Dec 16 '15 at 20:20
  • $\begingroup$ Block $A$ into a $1 \times 1$ block, a $1 \times (r-1)$ block, a $(r-1) \times 1$ block, and a $(r-1)\times(r-1)$ block (from top left to bottom right). Then use equation 2 here. This is a slight variant of what I suggested earlier, except you only need to invert the diagonal $(r-1) \times (r-1)$ block and a scalar. $\endgroup$ – JimmyK4542 Dec 16 '15 at 20:32
  • $\begingroup$ @JimmyK4542 Wow. What a very, very sleek formula. Please feel free to post that as an answer and I can award you points. Thank you very much. $\endgroup$ – Clarinetist Dec 16 '15 at 20:39
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First, partition the matrix into blocks $\begin{bmatrix}A & U \\ V & C\end{bmatrix}$, where:

$A = \displaystyle\sum_{i = 1}^{r}n_i$, $U = \begin{bmatrix}n_1 & n_2 & \cdots & n_{r-1}\end{bmatrix}$, $V = \begin{bmatrix}n_1 \\ n_2 \\ \vdots \\ n_{r-1}\end{bmatrix}$, and $C = \begin{bmatrix}n_1& & & \\ & n_2 & & \\ & & \ddots & \\ & & & n_{r-1}\end{bmatrix}$.

Then, using the block matrix inversion formula found in equation 2 here, we have:

$\begin{bmatrix}A & U \\ V & C\end{bmatrix}^{-1} = $ $\begin{bmatrix} (A-UC^{-1}V)^{-1} & -(A-UC^{-1}V)^{-1}UC^{-1} \\ -C^{-1}V(A-UC^{-1}V)^{-1} & C^{-1}V(A-UC^{-1}V)^{-1}UC^{-1}+C^{-1} \end{bmatrix}$.

Since $C$ is diagonal, we have $C^{-1} = \begin{bmatrix}n_1^{-1}& & & \\ & n_2^{-1} & & \\ & & \ddots & \\ & & & n_{r-1}^{-1}\end{bmatrix}$.

Then, we can easily compute $C^{-1}V = \begin{bmatrix}1\\1\\ \vdots \\ 1\end{bmatrix}$, $UC^{-1} = \begin{bmatrix}1&1& \cdots & 1\end{bmatrix}$, and $A-UC^{-1}V = \displaystyle\sum_{i = 1}^{r}n_i - \sum_{i = 1}^{r-1}n_i = n_r$, which gives us the following blocks:

$(A-UC^{-1}V)^{-1} = n_r^{-1}$

$-(A-UC^{-1}V)^{-1}UC^{-1} = \begin{bmatrix}-n_r^{-1} & -n_r^{-1} & \cdots & -n_r^{-1}\end{bmatrix}$

$-C^{-1}V(A-UC^{-1}V)^{-1} = \begin{bmatrix}-n_r^{-1} \\ -n_r^{-1} \\ \vdots \\ -n_r^{-1}\end{bmatrix}$

$C^{-1}V(A-UC^{-1}V)^{-1}UC^{-1}+C^{-1} = \begin{bmatrix}n_1^{-1}+n_r^{-1} & n_r^{-1} & \cdots & n_r^{-1} \\ n_r^{-1} & n_2^{-1}+n_r^{-1} & \cdots & n_r^{-1} \\ \vdots & \vdots & \ddots & \vdots \\ n_r^{-1} & n_r^{-1} & \cdots & n_{r-1}^{-1}+n_r^{-1}\end{bmatrix}$

Thus, the inverse of your bow-shaped matrix is:

$\begin{bmatrix}n_r^{-1} & -n_r^{-1} & -n_r^{-1} & \cdots & -n_r^{-1}\\ -n_r^{-1} & n_1^{-1}+n_r^{-1} & n_r^{-1} & \cdots & n_r^{-1} \\ -n_r^{-1} & n_r^{-1} & n_2^{-1}+n_r^{-1} & \cdots & n_r^{-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -n_r^{-1} & n_r^{-1} & n_r^{-1} & \cdots & n_{r-1}^{-1}+n_r^{-1}\end{bmatrix}$

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