We consider only the quadratic case. $f(x,y)=\frac{1}{2}x^TQx. $ And suppose we can choose $x_0$ to make $g_0$ in the span of its eigenvectors $e_i$, where $g_k=Qx_k$, being the gradient in each iteration.
I have already showed that each $g_k$ is also in the span of the eigenvectors. And in a special case where $g_0$ equals some eigenvector, it will converge in the next step. But for more general case, the calculation got messy.
Here I followed the theorem from "Nonlinear programming" Luenberger and Ye, 4th edition, here, section 8.6. The convergence rate is given by $1-\frac{(g_k^Tg_k)^2}{(g_k^TQg_k)(g_k^TQ^{-1}g_k))}$. When $g_0$ is an eigenvector, just by calculation, it is zero. But what about in general it is in the span of eigenvectors?
Any hints would be appreciated.