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I want to prove that the matrix product $A^{-1}B$ is positive definite where A is a symmetric positive definite matrix and B is a symmetric matrix. I have tried to use the following theorem but I did not come up with a good proof. Any good reference it would be appreciated also.

Theorem1:

If A is positive definite, then A is invertible and $A^{-1}$ is positive definite.

Thank you in advance!

EDIT: (added after Bertrand R answer)

Theorem2:

Let C be positive definite and D symmetric of the same order. Then there exist a non-singular matrix P and a diagonal matrix Λ such that $$ C = PP^{T} \, and \,D = PΛP^{T} $$

Theorem2 is extracted from the book "Matrix Differential Calculus with Applications in Statistics and Econometrics" and the proof is highly interesting. Also the first equality that refers to the positive definite matrix is also called the Cholesky decomposition.

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    $\begingroup$ Your hypotheses are probably incomplete. If $B=0$ (which is symmetric), then $A^{-1}B=0$ is not positive definite. $\endgroup$ – minar Aug 23 '13 at 12:29
  • $\begingroup$ If we suppose that B can not be zero can we prove what am I asking? $\endgroup$ – darkmoor Aug 23 '13 at 12:39
  • $\begingroup$ You got your answer. IMO, you should open a new question now, instead of just editing the old one. Also, I don't understand what your current question is, since you say that the above theorem has a "highly interesting proof" (so, you know it). $\endgroup$ – Vedran Šego Aug 24 '13 at 15:38
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I think you are refering to the following result :

if A is symmetric positive definite and B positive symmetric then AB is similar to a positive symmetric matrix.

Write $A=R^2$ where R is the "classic" square root of $A$ and is symmetric positive definite.

Then $$AB=R(RBR)R^{-1}$$ $RBR$ is symmetric, positive.

If you want to deal with $A^{-1}$ instead of $A$ do the same.

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  • $\begingroup$ From Cholesky-decomposition in A, R is an upper or lower triangular matrix. What theory did you use to conclude that RBR is positive and symmetric and how you finally mark AB as a symmetric and positive matrix? $\endgroup$ – darkmoor Aug 23 '13 at 13:45
  • $\begingroup$ Read more carefully please : _ Diagonalize A and take the square root of the diagonal to get R symmetric (e.g en.wikipedia.org/wiki/… ) _ R is then symmetric positive definite. _ Then I never said that AB was symmetric positive, but I said that is was similar to a symmetric positive matrix (which is $RBR$ in this case). $\endgroup$ – Bertrand R Aug 23 '13 at 14:00

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