How to find the inverse of the matrix $A = (a_{ij})$, where $a_{ij} = p^{|i-j|}$. Assuming the matrix is invertible.
What's the special thing of this matrix?
I can only know what this matrix is. But I don't know how to derive its inverse.
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1If $|p|<1$, then this post applies – Ben Grossmann Sep 28 '22 at 01:42
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It can be interesting to see my answer here where I give an interpretation with autoregressive processes. – Jean Marie Sep 28 '22 at 09:34
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This matrix is known as a Kac-Murdock-Szego (KMS) matrix. Apparently, its inverse has the nice formula described here whenever it is invertible (which occurs whenever $p \neq \pm 1$). In particular, we have $$ A^{-1} = \frac 1{1 - p^2}\pmatrix{ 1&-p&0&\cdots&0\\ -p & 1 + p^2 & -p & \ddots&\vdots\\ 0&-p&1+p^2&\ddots&0\\ \vdots&\ddots&\ddots&\ddots&-p\\ 0 & \cdots & 0 & -p & 1+p^2}. $$
Ben Grossmann
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1If you want a proof, it's easy to compute the entries of the product $AA^{-1}$ and show that they correspond to the entries of the identity matrix – Ben Grossmann Sep 28 '22 at 01:55
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@Jonathen You multiply this matrix with the original and check it gives you the identity. Since cols and rows are almost all zero it is easy to compute the dot product. – Nicolas Bourbaki Sep 28 '22 at 01:55
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@Jonathen If you really want a "derivation", then you might be content with applying this formula for the inverse of the $L$ from the Cholesky decomposition of $A$ as given in this post that I linked earlier. This technically only applies in the case that $|p|<1$, but you could argue that any formula that applies for all $p$ with $|p|<1$ must apply more generally. – Ben Grossmann Sep 28 '22 at 01:58