I have an exercise where I have to say if a vector $\vec{u} = \left(\begin{matrix}1 \\ 2\end{matrix}\right)$ is or not in a column space of a matrix $A = \left(\begin{matrix}1 & -2\\ -2 & 4\end{matrix}\right)$. I am quite newbie, and I am still not so comfortable with these concepts.
What I did was to create an augment matrix:
$$\left(\begin{array}{cc|c}1 & -2 & 1 \\ -2 & 4 & 2\end{array}\right)$$
If I try to reduce this I get the second row like $\left(\begin{array}{cc|c}0 & 0 & 4 \end{array}\right)$.
Which I think means that the $\vec{v}$ is not in $A$, because $0 +/- 0 \neq 4$.
Could you provide a more technical explanation of why? I know this might seem to easy, but just to understand better…
Best Answer
You could form the projection matrix, $P$ from matrix $A$:
$$P = A(A^TA)^{-1}A^T$$
If a vector $\vec{x}$ is in the column space of $A$, then
$$P\vec{x} = \vec{x}$$
i.e. the projection of $\vec{x}$ unto the column space of $A$ keeps $\vec{x}$ unchanged since $\vec{x}$ was already in the column space.
$\therefore$ check if $P\vec{u} = \vec{u}$