I'm looking for ways to solve the inverse of a symmetric matrix with 1's on the diagonal and all entries $x_{ij}$ continues variables within the range $(-1,1)$. The matrix thus looks like: \begin{bmatrix} 1 & x_{12} & x_{13} & \dots & x_{1m} \\ x_{12} & 1 & x_{23} & \dots & x_{2m} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{1m} & x_{2m} & x_{3m} & \dots & 1 \end{bmatrix}

Is there a way to solve this matrix analytically for any $m$? How should I go about to try and solve this?

Thanks a lot for any advice on this!


  • $\begingroup$ I reckon there will be nothing simpler than good old $\frac{1}{\det M}(Com(M))^T$. This is too close to the general case. $\endgroup$ – Arnaud Mortier Feb 25 '18 at 14:59
  • $\begingroup$ Also note that $\pmatrix {1 & a \\ a & 1}$ is invertible for $|a| < 1$, but for $|a| = 1$, it's not. So some of your matrices are very nearly not invertible. If you're hoping to find a useful formula and use a computer to evaluate it, this non-invertibility may be revealed in floating-point sensitivity: a tiny change to one entry makes a massive change to one or more entries in the inverse. Frankly, for computations: you're probably better off using the SVD and/or pseduoinverse. $\endgroup$ – John Hughes Feb 25 '18 at 15:32
  • $\begingroup$ Hmm yeah I know.. I was hoping there was some rule I was missing, because it's part of a greater problem for which I would like to have a nice expression for this problem. A few months ago, I was stuck on this same issue, but with a fixed $x$ and someone found a solution to simplify it very nicely. math.stackexchange.com/questions/2544652/… I guess I was hoping on finding something similar :) $\endgroup$ – blah_crusader Feb 25 '18 at 15:44
  • $\begingroup$ .This is a positive definite matrix. Maybe it can be helpful to you: Study Cholesky decomposition method and SVD method in matrix analysis $\endgroup$ – M.Bashiri Feb 25 '18 at 16:05

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