Positive definite and positive definite submatrices
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From: jimbospace on 27 Nov 2009 00:34 Here is my problem: 1. A is a symmetric positive definite (PD) matrix if and only if the determinant of all its principal submatrices (Ak) are positive. Right? 2. det |Ak| = Product of eigenvalues of Ak. Thus, if Ak is PD, then det|Ak| > 0. Can then I say that: If all Ak are PD => A is PD? But all I can find around is the opposite condition: A is PD => all Ak are PD I don't get it. Can someone help me pointing out what is wrong? Thanks,
From: Ken Pledger on 16 Dec 2009 15:00 In article <361752765.45368.1259336074246.JavaMail.root(a)gallium.mathforum.org>, jimbospace <jimbospace1(a)yahoo.fr> wrote: > Here is my problem: > > 1. A is a symmetric positive definite (PD) matrix if and only if the > determinant of all its principal submatrices (Ak) are positive. Right? Yes, if you include A itself as a submatrix. > .... > Can then I say that: > > If all Ak are PD => A is PD? > .... If A itself is one of the Ak, then this is trivial. But if you exclude A itself then you're in trouble. For example, (1 2) (2 1) is not positive definite. Ken Pledger.
From: Robert Israel on 16 Dec 2009 15:36
Ken Pledger <ken.pledger(a)mcs.vuw.ac.nz> writes: > In article > <361752765.45368.1259336074246.JavaMail.root(a)gallium.mathforum.org>, > jimbospace <jimbospace1(a)yahoo.fr> wrote: > > > Here is my problem: > > > > 1. A is a symmetric positive definite (PD) matrix if and only if the > > determinant of all its principal submatrices (Ak) are positive. Right? > > > Yes, if you include A itself as a submatrix. Assuming, of course, that A is real and symmetric. The determinants are not going to help you with that. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |