Solving simple quadratic function using conjugate gradient method - why are my directions linearly dependent?

Question

I have a simple function:

$$ f(x)=x_1^2 + 2 x_2^2 -2 x_1 x_2 - 2 x_2 + 2 x_1$$ And I'd like to solve it using conjugate gradient method at initial point $x^0=(2,2)$. I found the hessian

$$ Q= \begin{bmatrix} 1 & -1 \\ -1 & 2 \\ \end{bmatrix} $$ And I found the conjugate directions starting with $d^0=-\nabla{f}(x_0)=[-2,-2].$

Now the next conjugate direction using the Gram Schmidt procedure is $d^1=[0,0]$. Usually this means that Gram Schmidt terminates and those are the directions. But this doesn't seem correct.

Am I not doing this properly? It would be great to see how to get the conjugate directions, and also how to apply the method to this problem. (Conjugate gradient method should only take max 2 iterations for this problem in theory).

What do you want to "solve" ? Is it the values $(x,y)$ realizing the minimum of $f$ : $\operatorname{argmin} f(x,y)$ ? — Jean Marie, May 03 '23 at 18:36
@BobDobbs Not the gradient I'm getting at the initial point (2,2). Perhaps you evaluated it at (0,0)? — somewhere, May 04 '23 at 04:35
@JeanMarie Yes. By "solve" I mean obtain the minimizer (x1, x2). It should be possible by only taking 2 steps of the Conjugate Gradient method, since there are only 2 dimensions, and the Hessian is positive definite. — somewhere, May 04 '23 at 04:37
@somewhere Oh yes. They are linearly dependent. The algorithm stops. — Bob Dobbs, May 04 '23 at 09:48
Right. But using those directions doesn't give me the minimum value. — somewhere, May 04 '23 at 14:55
The minimizer is $[-2,0]$, and those directions don't result in that as the final point. — somewhere, May 04 '23 at 15:22

Solving simple quadratic function using conjugate gradient method - why are my directions linearly dependent?

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