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I have a square matrix with elements $A_{i,j} = b^{|i-j|}$ where $0 < b < 1$. I want to prove that it is positive definite. The matrix looks like this: $$ A_{n\times n}:=\left( \begin{array}{cccc} \;1 & b & b^2&\ldots& b^{|1-n|}\\ b & 1 & b &\ldots & b^{|2-n|}\\ b^2& b & 1 & \ldots& b^{|3-n|}\\ \vdots& \vdots & \vdots &\ddots\\ b^{|n-1|} & b^{|2-n|}& b^{|3-n|} &\ldots b^{|n-n|} \end{array}\right) $$ Can anyone show it is positive definite?


marked as duplicate by user1551, Dan Rust, Lord_Farin, Chris Godsil, mdp Oct 31 '13 at 12:19

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  • $\begingroup$ If you can get a formula for the determinant of the principal minors, and if you can prove those determinants are positive, you win. If you can find the eigenvalues, and prove that they are positive, you win. $\endgroup$ – Gerry Myerson Oct 31 '13 at 11:31
  • $\begingroup$ What's the dimension of this matrix? $\endgroup$ – Shuchang Oct 31 '13 at 11:37
  • $\begingroup$ Ah, you're right -- this is the same as the exponential kernel, and so the proof of that one works for this. $\endgroup$ – user2316794 Oct 31 '13 at 11:50
  • $\begingroup$ But this one can be answered without using the exponential kernel (see Davide Giraudo's answer, for instance), because the "data points" here have a very special structure, namely, they are simply the numbers $0,1,2,\ldots$. But still, I would prefer a proof using the exponential kernel, as it provides a more general perspective. $\endgroup$ – user1551 Oct 31 '13 at 12:25

We denote by $\Delta_n$ the determinant of $A_n$. Considering the row $n$, and replacing it by $R_n-bR_{n-1}$ (which does not change the determinant), we get the recursion relation $\Delta_n=(1-b^2)\Delta_{n-1}$. Since $\Delta_2=1-b^2$, the principal minors have a positive determinant. We conclude by Sylvester's criterion.


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