# Prove that matrix is positive definite [duplicate]

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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, mdpOct 31 '13 at 12:19

• 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. – 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.