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The symmetric matrix I need to invert is of the following form:

\begin{align} J_e=\left(\begin{matrix}-I&B_{11}&B_{12}&...&...&B_{1(N-1)}\\ B_{11}&-I&B_{22}&...&...&B_{2(N-1)}\\ \vdots&&\ddots&&&\vdots\\ B_{i1}&&&-I&...&B_{i(N-1)}\\ \vdots&...&B_{ij}&...&\ddots&\vdots\\ B_{(N-1)1}&...&...&...&...&-I\end{matrix}\right) \end{align}

where $I=I_{N-1}$ for some $N\geq3$ and $B_{ij}=B_{ji}'$ are $(N-1)\times(N-1)$ matrices with only one non-zero entry $[B_{ij}]_{ij}\neq0$, thus they all have rank one. The entire matrix is thus $N(N-1)\times N(N-1)$.

My question: How can I analytically obtain $J_e^{-1}$ exploiting the above structure?

I was thinking about writing it as \begin{align} J_e=M-I=\left(\sum_{k=1}^{N-1}M_k\right)-I \end{align}

where each $M_k$ is a monomial matrix, but the requirements of the results I found so far that relate to this are always to demanding for what the above problem.

Any hint or idea is very much appreciated.

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  • $\begingroup$ Just to confirm, the nonzero entry of $B_{ij}$ is the $(i,j)$ entry? Then you could permute rows and columns so that the only nonzero off-diagonal blocks are $M,M^T$, where $M$ is the upper triangular matrix with entries $M_{ij}=[B_{ij}]_{ij}$. Then use the formula for inversion of a 2x2 block matrix. $\endgroup$
    – Nick Cook
    Sep 13, 2016 at 17:54
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    $\begingroup$ @Nick: It seems to me that after a permutation, the onl nonzero off-diagonal elements are in a $N\times N$ matrix with diagonal elements equal to $-1$ and off-diagonal elements $[B_{ij}]_{ij}$. So the task is equivalent to inverting a symmetrical $N\times N$ matrix. $\endgroup$
    – Ilya Bogdanov
    Sep 13, 2016 at 20:04
  • $\begingroup$ Ah, yes, that is a different permutation that puts it in a slightly more compact form. I was thinking of the one that puts $M$ in the position initially occupied by $B_{11}$, say, so then you reduce to inverting the $2(N-1)\times 2(N-1)$ symmetric matrix $\begin{pmatrix} -I & M\\ M^T & -I\end{pmatrix}$. Both should give the same answer :). $\endgroup$
    – Nick Cook
    Sep 14, 2016 at 4:43

2 Answers 2

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Let $P$ be an $N(N-1)\times N(N-1)$ permutation matrix such that $Pe_i=e_{iN-(N-1)}$ and $Pe_{N-1+j}=e_{jN}$ for each for each $1\le i,j\le N-1$, where $e_i$ is the $i$th standard basis vector. Letting $M$ denote the $N-1\times N-1$ upper triangular matrix with entries $M_{ij} = [B_{ij}]_{ij}$, we have $$ P^T J_e P = \begin{pmatrix} -I_{N-1} & M & 0\\ M^T & -I_{N-1} & 0 \\ 0 &0 & -I_{(N-2)(N-1)} \end{pmatrix} $$ where $I_k$ denotes the $k\times k$ identity matrix. The top-left $2(N-1)\times 2(N-1)$ submatrix can be inverted using the Schur complement formula (https://en.wikipedia.org/wiki/Schur_complement), giving $$ J_e^{-1} = P \begin{pmatrix} (MM^T-I_{N-1})^{-1} & (MM^T-I_{N-1})^{-1}M &0\\ M^T(MM^T-I_{N-1})^{-1} & (MM^T-I_{N-1})^{-1} & 0 \\ 0 &0 & -I_{(N-2)(N-1)} \end{pmatrix} P^T. $$ As @Ilya commented above, one could also choose a permutation that puts all nontrivial entries in the top-left $N\times N$ submatrix.

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I doubt that there's a simple closed-form formula. In principle you could get it for a particular $N$ by repeated use of the Sherman-Morrison formula, but you might as well just construct the matrix and invert it explicitly with a CAS.

For the case $N=3$ with $(B_{ij})_{ij} = b_{ij}$ I get $J^{-1} = \det(J)^{-1} \text{adj}(J)$ with $\det(J) ={b_{{1,1}}}^{2}{b_{{2,2}}}^{2}-{b_{{1,1}}}^{2}-{b_{{1,2}}}^{2}-{b_{{2, 2}}}^{2}+1$ and $$ \text{adj}(J) = \left[ \begin {array}{cccccc} {b_{{2,2}}}^{2}-1&0&b_{{1,1}}{b_{{2,2}} }^{2}-b_{{1,1}}&-b_{{2,2}}b_{{1,2}}&0&-b_{{1,2}}\\ 0 &-{b_{{1,1}}}^{2}{b_{{2,2}}}^{2}+{b_{{1,1}}}^{2}+{b_{{1,2}}}^{2}+{b_{{ 2,2}}}^{2}-1&0&0&0&0\\ b_{{1,1}}{b_{{2,2}}}^{2}-b_{{ 1,1}}&0&{b_{{1,2}}}^{2}+{b_{{2,2}}}^{2}-1&-b_{{2,2}}b_{{1,1}}b_{{1,2}} &0&-b_{{1,1}}b_{{1,2}}\\ -b_{{2,2}}b_{{1,2}}&0&-b_{{ 2,2}}b_{{1,1}}b_{{1,2}}&{b_{{1,1}}}^{2}+{b_{{1,2}}}^{2}-1&0&{b_{{1,1}} }^{2}b_{{2,2}}-b_{{2,2}}\\ 0&0&0&0&-{b_{{1,1}}}^{2}{ b_{{2,2}}}^{2}+{b_{{1,1}}}^{2}+{b_{{1,2}}}^{2}+{b_{{2,2}}}^{2}-1&0 \\ -b_{{1,2}}&0&-b_{{1,1}}b_{{1,2}}&{b_{{1,1}}}^{2}b _{{2,2}}-b_{{2,2}}&0&{b_{{1,1}}}^{2}-1\end {array} \right] $$ For $N=4$, I'll just write the determinant: $$\eqalign{& -{b_{{1,1}}}^{2}{b_{{2,2}}}^{2}{b_{{3,3}}}^{2}+{b_{{1,1}}}^{2}{b_{{2,2 }}}^{2}+{b_{{1,1}}}^{2}{b_{{2,3}}}^{2}+{b_{{1,1}}}^{2}{b_{{3,3}}}^{2}+ {b_{{1,2}}}^{2}{b_{{2,3}}}^{2}+{b_{{1,2}}}^{2}{b_{{3,3}}}^{2}\cr&-2\,b_{{1 ,2}}b_{{1,3}}b_{{2,2}}b_{{2,3}}+{b_{{1,3}}}^{2}{b_{{2,2}}}^{2}+{b_{{2, 2}}}^{2}{b_{{3,3}}}^{2}-{b_{{1,1}}}^{2}-{b_{{1,2}}}^{2}-{b_{{1,3}}}^{2 }-{b_{{2,2}}}^{2}-{b_{{2,3}}}^{2}-{b_{{3,3}}}^{2}+1} $$

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