if A is positive definite, then Ax > B eq to x > A^-1 B
... Can you prove that?
Note what I said, for equation: if a = b then c = d <=> a+b=c+d
but for inequation, if a > b then c > d => a+b > c+d but a+b > c+d (not =>) c > d
A simple counterexample:
A = [ 1 1 ; 1 2] is PD B = [2;3] and
AX > B (1)
But we can't say
X > A^-1 * b = [1;1] (2)
Since obviously x1 = 0, x2 = 3 fits (1) but not (2)
but for inequation, if a > b then c > d => a+b > c+d but a+b > c+d (not =>) c > d
A simple counterexample:
A = [ 1 1 ; 1 2] is PD B = [2;3] and
AX > B (1)
But we can't say
X > A^-1 * b = [1;1] (2)
Since obviously x1 = 0, x2 = 3 fits (1) but not (2)
Yeah!
Huasing Association 1999 - 2013