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Let $$ A= \begin{bmatrix} n & n_1 & n_2 & \cdots & n_s & 0 \\ n_1 & n_1 & 0 & \cdots & 0 & 1 \\ n_2 & 0 & n_2 & \cdots & 0 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n_s & 0 & 0 & \cdots & n_s & 1 \\ 0 & 1 & 1 & \cdots & 1 & 0 \end{bmatrix}, $$ where $n=\sum_{i=1}^sn_i$.

My questions: What is $A^{-1}$? Does it have an elegant expression?

My attempts: I tried some small matrices and found that $A^{-1}$ has the following form: $$ A^{-1}=\begin{bmatrix} \cdots & \cdots & \cdots & \cdots & \cdots & -\frac{1}{s} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\ -\frac{1}{s} & \frac{1}{s} & \frac{1}{s} & \cdots & \frac{1}{s} & 0 \end{bmatrix} $$

While I faild to find patterns in those "dots" parts.

Thanks a lot.

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  • $\begingroup$ If all the $n_i$'s are $0$, the matrix is not invertible. $\endgroup$ Commented Sep 1, 2020 at 9:43
  • $\begingroup$ @TheSilverDoe, thanks for the comment. But my interest lies in when $n_i$ are all positive numbers, or more specifically, positive integers. I suppose in this case, $A$ is invertible? $\endgroup$ Commented Sep 1, 2020 at 9:57
  • $\begingroup$ I suspect that a nice approach would be to compute the inverse of the submatrix $$ \begin{bmatrix} n & n_1 & n_2 & \cdots & n_s \\ n_1 & n_1 & 0 & \cdots & 0 \\ n_2 & 0 & n_2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n_s & 0 & 0 & \cdots & n_s \end{bmatrix}, $$ then compute the full inverse using the Woodbury matrix identity $\endgroup$ Commented Sep 1, 2020 at 10:32
  • $\begingroup$ Actually, it turns out that this submatrix is never invertible, which makes my suggestion problematic $\endgroup$ Commented Sep 1, 2020 at 10:38

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Partial Answer: Here's a strategy you might find helpful.

Let $M$ denote the submatrix of $A$ given by $$ M = \pmatrix{n & n_1 & n_2 & \cdots & n_s \\ n_1 & n_1 & 0 & \cdots & 0 \\ n_2 & 0 & n_2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n_s & 0 & 0 & \cdots & n_s}. $$ we see that $M$ can be expressed in the form $$ M = \sum_{i=1}^s (e_1 + e_{1+i})(e_1 + e_{1+i})^T, $$ where $e_1,e_2,\dots,e_{s+1}$ denotes the canonical basis of $\Bbb R^n$. This gives us the decomposition $M = BDB^T$, where $$ B = \pmatrix{e_1 + e_2 & e_1 + e_3 & \cdots & e_1 + e_{s+1}} = \pmatrix{1&1&\cdots&1\\1\\&1\\ && \ddots \\&&&1}, \quad D = \pmatrix{n_1 \\ & \ddots \\ && n_s}. $$ So, the matrix $A$ can be written in the form $$ A = \pmatrix{BDB^T & x\\ x^T & 0}, $$ where $x = (0,1,1,\dots,1)^T$.

Let $C$ denote the matrix $$ C = \pmatrix{1&-1&\cdots & -1\\ &1\\ &&\ddots\\ &&&1}. $$ We note that $$ CB = \pmatrix{0\\&I_{s}}. $$ With that in mind, we find that $$ \tilde C A \tilde C ^T = \overbrace{\pmatrix{C&0\\0 & 1}}^{\tilde C} \pmatrix{BDB^T & x\\ x^T & 0} \pmatrix{C&0\\0 & 1}^T = \pmatrix{D & Cx\\ (Cx)^T & 0}. $$ It would suffice to find the inverse of this "nicer" matrix $\tilde C A \tilde C ^T$, then compute $$ A^{-1} = \tilde C^T[\tilde C A \tilde C ^T]^{-1}\tilde C. $$

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  • $\begingroup$ Thanks for the hint, which is very helpful! But I suppose there are a few typos? For example, $B$ should be a $s+1$ by $s$ matrix, here column $e_1+e_{s+1}$ is missing. Besides, $CB$ should be $s+1$ by $s$ matrix with a zero vector as its first row and $I_s$ as the other part. As a result, the upper left corner of $\tilde{C}A\tilde{C}^T$ should be a $s+1$ by $s+1$ diagonal matrix with $0, n_1, \dots, n_s$ as its diagonal elements. $\endgroup$ Commented Sep 2, 2020 at 8:51
  • $\begingroup$ @Chris Yes, that's all correct. I wasn't too careful when typing this up; sorry for any confusion $\endgroup$ Commented Sep 2, 2020 at 9:43

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