The problem comes from my textbook:
Let
$$ f(x)=\dfrac{1}{2}x^TAx-x^Tb+c $$ Q: if the steepest descent method is used on this problem, what happens if the starting value $x_0$ is such that $x_0 - x^*$ is an eigenvector of $A$ (symmetric and positive definite), where $x^*$ is the solution?
Here's what I've tried:
The steepest descent direction: $p_k=-Ax_k+b$
The step size
$$t_k=\dfrac{p_k^Tp_k}{p_k^TAp_k}=\dfrac{(b-Ax_k)^T(b-Ax_k)}{(b-Ax_k)^TA(b-Ax_k)}$$
As $x_0-x^*$ is an eigenvector of $A$, we have
$A(x_0-x^*)=\lambda(x_0-x^*),\quad\nabla f(x^*)=Ax^*-b=0$
therefore $p_0=-Ax_0+v=\lambda(x^*-x_0)$, $t_0=\dfrac{1}{\lambda}>0$.
The question seems to hint me that some specific phenomenon appears when $x_0$ is taken such initial value. But through the calculation above, I note nothing strange. I cannot figure out the implication of the question, could someone enlighten me please?