Minimising problem with sums and matrices

Question

We assume that $p_0,...,p_k$ are pairwise conjugate to Q and we define a function $f(x)=\frac{1}{2}x^TQx+g^Tx$, and then we have to show that the solution $\alpha_0,...,\alpha_k$ is $\alpha_k=-\frac{x_k^Tp_k+g^Tp_k}{p_k^TQp_k}$, is solution to following minimisation problem $min_{\alpha_0,...,\alpha_k}f(\sum_{i=0}^k \alpha_ip_i)$. So I end up with finding the the minimum of $\alpha$ for $$min_{\alpha_0,...,\alpha_k}f(\sum_{i=0}^k \alpha_ip_i)=min_{\alpha_0,...,\alpha_k}(\frac{1}{2}(\sum_{i=0}^k \alpha_ip_i)^TQ(\sum_{i=0}^k \alpha_ip_i)+g^T(\sum_{i=0}^k \alpha_ip_i))$$But how do I show the minimum is gives as the given expression, I think some matrix computation tricks, but can see how to start. Can anyone help me with some tricks?

Answer 1

The classic conjugate gradient method. So you want to find $\alpha_{k}$ such that $f\left(x_{k}+\alpha_{k}p_{k}\right)$ is minimum.

$$ \begin{aligned} f\left(x_{k}+\alpha_{k}p_{k}\right) &= \frac{1}{2}x_{k}Qx_{k}+\alpha_{k}p_{k}^{T}Qx_{k}+\frac{1}{2}\alpha_{k}^{2}p_{k}^{T}Qp_{k}+q^{T}x_{k}+\alpha q^{T}p_{k} \\\\ 0 &= \frac{d}{d\alpha_{k}} f\left(x_{k}+\alpha_{k}p_{k}\right) \\ &= p_{k}^{T}Qx_{k} + \alpha_{k}p_{k}^{T}Qp_{k} + q^{T}p_{k} \end{aligned} $$

So there is a mistake in your post. It should be $\alpha_{k}=-\frac{p_{k}^{T}Qx_{k}+q^{T}p_{k}}{p_{k}^{T}Qp_{k}}$