0
$\begingroup$

I came across this problem when proving another proposition: I have $A=D_1+D_2P$ where $P$ is a symmetric, real-valued, positive definite matrix and $D_1$ and $D_2$ are diagonal matrices with positive, nonzero diagonal elements. I need to prove that $A$ is invertible.

My arguement on this problem is as follows: Since $D_2$ is diagonal with positive, nonzero elements, the determinant of the i'th principal minor of $D_2P$ is $$d_{21}...d_{2i}\left | P_i \right |>0$$ where $\left | P_i \right|$ is the determinant of i'th principal minor of $P$ which is positive since $P$ is positive definite.

Since all principal minors of $D_2P$ have positive determinants, I deduce that $D_2P$ is positive definite. Also it's clear that $D_1$ is positive definite; hence, $A$ is positive definite and invertibe.

I believe that my proof is wrong, specially since $D_2P$ needs not to be symmetric and I'm not sure the result on determinants of principal minors for positive definite matrices could be generalized to nonsymmetric cases.

I'd appreciate your comments on my proof, or If you could propose another way to prove this problem.

$\endgroup$
1
$\begingroup$

The following proof works just fine (using only the similarity transformation and the fact that the sum of two SPD matrices is SPD):

The matrix $A$ has the same eigenvalues as $$ B=D_2^{-1/2}AD_2^{1/2}=D_1+Q, $$ where $Q=D_2^{1/2}PD_2^{1/2}$. Hence $A$ is nonsingular iff $B$ is nonsingular. Since $P$ is SPD, $Q$ is SPD as well. The matrix $B$ is a sum two SPD matrices (one of them diagonal) and hence is SPD as well and therefore nonsingular.

$\endgroup$
  • $\begingroup$ nice and simple proof, thanks :) $\endgroup$ – Iman E Sep 30 '14 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.