To find a answer, I will use CG and its variants (BiCG/FLR) to solve some linear systems $Hx = b$ with respect to different $H$ and $b$.
The CG(Conjugate gradient method) is
function [x] = conjgrad(A,b,x)
r = b - A*x;
p = r;
rsold = r'*r;
for i = 1:length(b)
Ap = A*p;
alpha = rsold/(p'*Ap);
x = x + alpha*p;
r = r - alpha*Ap;
rsnew = r'*r;
if sqrt(rsnew) < 1e-10
break;
end
p = r + (rsnew/rsold)*p;
rsold = rsnew;
end
end
The following examples are designed:
- Let $H=diag(1,-1)$, $b = (1, 2)$, and $x_0=(0,0)$, CG converges in two steps.
- Let $H=diag(1,-1)$, $b = (1, 1)$, and $x_0=(0,0)$, CG breaks down in the first step, since $p^TAp = 0$
Conclusion: The convergece of CG method cannot be used to identify if $H$ is positive definite.
- Let $H=diag(1,-1)$, $b = (1, 2)$, and $x_0=(0,0)$, then $p^TAp <0$ in second step
- Let $H=diag(1,-1)$, $b = (1, 0)$, and $x_0=(0,0)$, CG converges in the first step, and $p^TAp = 1 >0$ in the first step, therefore no $p^TAp<0$ is encountered.
Conclusion: The value $p^TAp$ is not sufficient to show if $H$ is positive definite.
