I was studying Newton's method recently, and I was trying to get a step-size with exact line search for a quadratic problem,e.g.$f(X) = \frac{1}{2}X^TQX$. I have used the same way as what has been done with gradient descent but replace the descent direction as $h=(\nabla ^{2} f(x))^{-1}\nabla f(x)$ in $\alpha = \frac{h^T QX}{h^TQh}$, but I found it turned out that the step-size is always 1, which seems to be nonsense. Can someone tell me how to fix it?
P.S. $f:\{\rm I\!R^n\rightarrow \rm I\!R \}$,$Q \in \rm I\!R ^{nxn}$.