Spectral radius of non-negative matrix

Hi Vikram,

>I think that you have given the right way to solve my problem for
>addressing my research.

Great! Would you mind sharing in which application this problem appears?


>Two final queries:
>
>a). In Wainwright's theorem, the off-diagonal entries in the X matrix
>are the same (equaling B). This requirement is not necessary for the S-
>Lemma, right? That applies to my case.

Now, looking back, it seems that S-lemma requires symmetric (or hermitian)
matrices. So (I-G), and hence G, needs to be hermitian to invoke it.
However, if your G is not hermitian, the story is not over. There is a neat
trick to work around it: define H=(G+G*)/2, which implies
   (I-G)>0 => (I-H)>0 and   (I+G)>0 => (I+H)>0
and then invoke S-lemma with this matrix. The result should look like
   1-(v1+v2)'/2 * [I-(F+F*)/2]^{-1} * (v1+v2)/2 >0
and
   1-(v1+v2)'/2 * [I+(F+F*)/2]^{-1} * (v1+v2)/2 >0.


>b). It looks like your proof to show that Spectral Radius(G)<1 implies
>I+G and I-G are positive definite is also a necessary condition.

Do you mean |G|<1 => (I+G)>0 and (I-G)>0?  If so, then yes.  I believe the
converse is also true.


>Thank you very much. It was a pleasure interacting with you.

Likewise. Good luck with your research.

Emre