thanks. now I see why I couldn’t prove it. I will have to think if I really need it.

]]>It’s false. Here is a counterexample in MATLAB:

>> rng(6); n=2; A = rand(n)/(1.1*max(abs(eig(A)))); (eye(n)+A)\A

ans =

2.3058e-01 2.3561e-01

9.5246e-02 -1.3619e-02

With a change of sign, (I-Q)^{-1}Q is nonnegative, as can be seen

from a Neumann expansion of the inverse.

For square non negative Q with spectral radius < 1, (I hope that)

(I + Q)^{-1}Q is also non negative.

Any hints or thoughts?

]]>I was fortunate to be in Jack’s ‘Perfeccionamiento’ Spanish group at the Instituto Cervantes during 2010-13 approximately. That group was fantastic, largely due to Jack’s vigorous contributions! We would go to the pub round the corner afterwards and discuss and debate topical issues. I was aware that Jack was a man of stature and accomplishment but he never let on how important his contribution to mathematics in academia had been.

]]>This blog says ” the cholesky is guaranteed to run to completion if the condition number k2(A) is safely less than u^-1″. But you have used Wilkinson’s bound in other articles and books, namely 20n^(3/2)*k2(A)*u^-1 < 1. Did I miss something? ]]>

I might not have commented before, but I have been responsible for some of those 1e6 views 🙂

]]>Yes, x doesn’t have to be from the normal distribution.

]]>