Let $A$ be a symmetric positive definite matrix and fix $b \in \mathbb{R}^n$. Consider solving for the solution $x^* = A^{-1}b$ by using a projection method. In other words, we have a sequence of approximate solutions $\{x_k\}_k$ with some initial guess $x_0$. In order to get $x_{k+1}$ from $x_k$ we use an orthogonal projection method on the subspaces $K_k = L_k = \text{span}\{r_k, Ar_k\}$, where $r_k = b-Ax_k$.
Essentially the projection method can be described as follows, given $x_k$ we have $$x_{k+1} = x_k + \delta, \quad \delta \in K_k \\ r_{k+1} = b-Ax_{k+1} \perp K_k.$$
Now this problem comes from Saad: Define the vector $p$ to be the orthogonalization of $Ar$ with respect to $r$. Find a formula for $p$. From here, use the formula for $p$ to write down the algorithm of the projection method above.
Not clear on the wording "$p$ is the orthogonalization of $Ar$ with respect to $r$", but if I had to guess they probably mean that $$p = Ar - \frac{\langle Ar, r \rangle}{\langle r, r \rangle} r,$$ so that $p \perp r$. But from here I am not sure how we can even use $p$. In general what we would do is write $x_{k+1} = x_k + c_1 r_k + c_2 Ar_k$ and use the condition that $r_{k+1} \perp K_k$ to deduce the constant $c_1$ and $c_2$, however doing this is very messy.
On the other hand, one could set $x_{k+1} = x_k+cp_k$ since $p_k \in K_k$, but then the condition $r_{k+1} \perp K_k$ gives two different values for $c$.
Any guidance is appreciative.
