[math-fun] A linear algebra theorem?
Is this a theorem? Let A and B be real symmetric matrices all of whose eigenvalues are positive. Then all eigenvalues of AB are positive. -- Gene
http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product implies that the product of two Hermitian matrices is diagonalisable with real eigenvalues of the same signs as the second factor, and gives Prop 6.1 of Denis Serre's _Matrices_ as a reference. -Thomas C On Mon, Mar 28, 2011 at 3:42 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Is this a theorem?
Let A and B be real symmetric matrices all of whose eigenvalues are positive. Then all eigenvalues of AB are positive.
-- Gene
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In simplest terms, the product of two positive definite-matrices is positive-definite. See http://en.wikipedia.org/wiki/Positive-definite_matrix -Robert On Mon, Mar 28, 2011 at 3:15 PM, Thomas Colthurst <thomaswc@gmail.com> wrote:
http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product implies that the product of two Hermitian matrices is diagonalisable with real eigenvalues of the same signs as the second factor, and gives Prop 6.1 of Denis Serre's _Matrices_ as a reference.
-Thomas C
On Mon, Mar 28, 2011 at 3:42 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Is this a theorem?
Let A and B be real symmetric matrices all of whose eigenvalues are positive. Then all eigenvalues of AB are positive.
-- Gene
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I'm not seeing where it says that; in fact, the article you cite seems to go out of its way *not* to say that. It says that if M and N are positive definite, then MNM and NMN are also, and that if M and N commute, then MN is positive definite. Surely if the general theorem were true, they would have said it, rather than presenting these weaker results. On Mon, Mar 28, 2011 at 4:17 PM, quad <quad@symbo1ics.com> wrote:
In simplest terms, the product of two positive definite-matrices is positive-definite.
See http://en.wikipedia.org/wiki/Positive-definite_matrix
-Robert
On Mon, Mar 28, 2011 at 3:15 PM, Thomas Colthurst <thomaswc@gmail.com> wrote:
http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product implies that the product of two Hermitian matrices is diagonalisable with real eigenvalues of the same signs as the second factor, and gives Prop 6.1 of Denis Serre's _Matrices_ as a reference.
-Thomas C
On Mon, Mar 28, 2011 at 3:42 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Is this a theorem?
Let A and B be real symmetric matrices all of whose eigenvalues are positive. Then all eigenvalues of AB are positive.
-- Gene
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On Mon, Mar 28, 2011 at 3:15 PM, Thomas Colthurst <thomaswc@gmail.com> wrote:
http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product implies that the product of two Hermitian matrices is diagonalisable with real eigenvalues of the same signs as the second factor, and gives Prop 6.1 of Denis Serre's _Matrices_ as a reference.
-Thomas C
Consider the product of the negative unit matrix (which is negative definite) and the unit matrix (positive definite) (of the same size, of course). Both are Hermitian. Would not that mean, that the product should be positive definite? Maybe I am missing something? Christoph
On Mon, Mar 28, 2011 at 3:42 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Is this a theorem?
Let A and B be real symmetric matrices all of whose eigenvalues are positive. Then all eigenvalues of AB are positive.
-- Gene
Is this a theorem? Let A and B be real symmetric matrices all of whose eigenvalues are positive. Then all eigenvalues of AB are positive. -- Gene ______________________________________________ A few remarks: 1. The product AB need not be symmetric. For nonsymmetric matrices, positive eigenvalues is stronger than positive-definite. The matrix [[1,1],[-1,1]] is positive-definite, but has eigenvalues 1 +- i. I'm looking for the stronger conclusion of positive eigenvalues. 2. If A and B are only required to be positive-definite, and not necessarily symmetric, then AB need not be positive-definite. For a counterexample, let A = B be a rotation in the plane by 60 degrees. 3. If A and B are only required to be real symmetric, then AB need not have real eigenvalues. Example: A = [[1,0],[0,-1]], B = [[0,1],[1,0]], AB = [[0,1],[-1,0]]. -- Gene
participants (5)
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Allan Wechsler -
Eugene Salamin -
Pacher Christoph -
quad -
Thomas Colthurst