Re: Matrix inverse question

I see how to show K(I - cK)^-1 by another continuity argument (if K
is invertible, we can move it inside the inverse and the resulting
thing is obviously spd, for small enough c. But you had a simpler
argument in mind using the fact that K and I-cK commute?

Thanks a lot!
David

On May 29, 7:59 pm, Chip Eastham <hardm...@xxxxxxxxx> wrote:
On May 29, 9:59 pm, David <david.dav...@xxxxxxxxx> wrote:

Hi,

Say I is the identity matrix, and M and K are real, symmetric positive
semi-definite (psd) matrices. All are n x n matrices, to be concrete.

Is I+MK necessarily invertible?

If MK were PSD (e.g. if M and K had t same eigenvetors), then it's
obviously true, but MK is not even symmetric in general.

Thanks,
David

Suppose M were actually positive definite, i.e.
invertible.  Then I + MK is not invertible implies
there exists nonzero vector u s.t. MKu = -u, and
thus Ku = -M^(-1)u.  But this is a contradiction
since u'Ku >= 0 while u'(-M^(-1)u) < 0.

So the conclusion would be that I + MK is invertible.
We can extend this to the case M is semi-definite
by a perturbation argument.  In particular:

I + MK = I + (M + cI)K - cK

and for sufficiently small c > 0, M + cI becomes
positive definite, I - cK is positive definite,
and:

I + (M + cI)K - cK = (I - cK) + (M + cI)K
= ( I + (M + cI)K(I - cK)^-1 ) (I - cK)

So it is only necessary to note that because K
commutes with I - cK, the product K(I - cK)^-1
must also be symmetric positive semi-definite.
Thus the general case reduces to the special
case M positive definite argued above.

regards, chip

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