How to determine an infinitely dimensional matrix is positive definite

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SUMMARY

This discussion addresses the determination of positive definiteness for infinitely dimensional matrices. It emphasizes that checking principal minors, a method applicable to finite-dimensional matrices, is not suitable for infinite dimensions. Instead, the discussion suggests verifying the condition > 0 for all non-zero vectors x in the space, which is a definition of positive definiteness in the context of inner products. Additionally, it highlights the importance of establishing that the operator is self-adjoint and that all eigenvalues must be positive to conclude positive definiteness.

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kevchang
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I ran into an infinitely dimensional matrix and wanted to show it is positive definite. I think I cannot do so by checking the principal minors as for finitely dimensional matrices... Can someone let me know how to show its positive definiteness? Thanks a ton!
 
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If you've got an inner product in that vector space, maybe you can check whether <Mx, x> > 0 for all x in the space. This implies that a matrix is positive definite in a finite dimensional space (if by positive definite you mean hermitian with non-negative eigenvalues). Not sure if it works in an infinite dimensional space.
 
Positive definite is defined using an inner product, so the post above should be taken as a definition, with the proviso that <Mx,x> > 0 for all non-zero x. The usual way to prove such things is to characterize the eigenvalues. In infinite dimensions, things get quite complicated, so one usually starts by verifying that the operator (matrix) is self-adjoint. Then a theorem states that there's an orthonormal basis of eigenvectors, so positive-definite reduces to showing that all eigenvalues are positive.
 

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