Re: Matrix problem


Sarah Moore schrieb:

Any real matrix A is monotone iff A*x > 0
implies x > 0 and A^-1 exists and is greater than zero.

But then it's quite trivial that if A and B have this property, then so
does AB.
You know that
A*x > 0 implies x>0
A is invertible
A^(-1) is greater than zero
B*x > 0 implies x>0
B is invertible
B^(-1) is greater than zero

You have to show that
A*B*x > 0 implies x>0
A*B is invertible
(A*B)^(-1) is greater than zero.

Hint: (A*B)^(-1) = B^(-1) * A^(-1)

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