I'm can't seem to figure out what this question is asking me:
Let
$$ A= \begin{bmatrix} -1 & 2 & 0 &0 \\ -1 & -1 & 1 & 0 \\ 0 & 1 & 1 &1 \\ 0 &0 & 2 &1\\ \end{bmatrix} r^0= \begin{bmatrix} 2\\ 1\\ -1\\ 0\\ \end{bmatrix} $$
Find the vector y that minimises $\lVert r^0-AV^1y\rVert_2$.
I've found both the Arnoldi decomposition in the form $AV^1=V^2H$, and the QR decomposition which are as follows:
$AV^1=V^2H$: $$ A \begin{bmatrix} \frac{\sqrt 6}{3}\\ \frac{\sqrt 6}{6}\\ -\frac{\sqrt 6}{6}\\ 0\\ \end{bmatrix} = \begin{bmatrix} \frac{\sqrt 6}{3} & \frac{2\sqrt 39}{39}\\ \frac{\sqrt 6}{6} & -\frac{5\sqrt 39}{39}\\ -\frac{\sqrt 6}{6} & -\frac{\sqrt 39}{39}\\ 0 & -\frac{\sqrt 39}{13}\\ \end{bmatrix} \begin{bmatrix} -\frac{2}{3}\\ \frac{\sqrt 26}{3}\end{bmatrix} $$ $QR$ decomposition :
$H=\begin{bmatrix} \frac{\sqrt 30}{15}\\ \frac{\sqrt 195}{15}\end{bmatrix} \begin{bmatrix} \frac{\sqrt 30}{3}\end{bmatrix}$
So I know what $r^0$, $A$, $V^1$ are, but what does y equal, and how do I minimise $\lVert r^0-AV^1y\rVert_2$ ?