Let $V$ be a collectino of the search direction for the conjugate gradient applied on a quadractic minimisation problem. As a proof of orthogonality in conjugate gradient:
$$ V^T V = I $$
Now supposed I have a vector $x$, and I want to express this as a sum of vectors that are parallel and perpendicular to the Krylov subspace spanned by the Hessian of this optimisation problem. From one of the article, this is quoted:
$$ x_{||} = V V^T x $$
I can't really work out the projection being used here and why that would work. Also, what is $V V^T$ if $V^T V$ is already an identity matrix?
You can write $x=y+z$, where $y=VV^Tx$ and $z=(I-VV^T)x$.
Notice that $V^Tz=0$ and so $z$ is orthogonal to the columns of $V$, while of course $y$ is generated by the columns of $V$.