# Is the perpendicular vector in a projection part of the nullspace?

Say I have a vector $\vec{b}$ that projects onto the column space of $\vec{a}$ to create a projected vector $\vec{p}$. This would also result in a perpendicular vector, which I call it $\vec{e}=\vec{b}-\vec{p}$. I used to understand that the vector $\vec{e}$ is in the nullspace of $N(\vec{a}^{T})$ because the dot product of the vector $\vec{e}\cdot\vec{p}=0$.

Recently, I thought that I could have misinterpreted the whole thing because I suddenly thought that $\vec{e}$ could have not been in the nullspace of $N(\vec{a}^{T})$ at all. I drew a sketch of a 2 dimension graph to visualise this thing but I think it applies to an extension of higher dimensions. In the picture below, it seems that $\vec{e}$ is only in the nullspace if it is $\vec{e^{'}}$ which I drew it in pink along the nullspace - that's $\vec{e}$ being moved all the way back.

So in this case, yes, any perpendicular vector along $\vec{a}$ would result in a dot product of zero scalar value. But how is $\vec{e}$ in the nullspace?

If $\vec{e}$ is not in the nullspace, then why does the projection formula $A^{T}\vec{e} =\vec{0} \Rightarrow A^{T}(\vec{b}-\vec{p})=\vec{0}$ works by really returning zero? It wouldn't have returned zero if $\vec{e}$ wasn't in the nullspace, would it?

Thanks for any help on this!

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Could you define $A$ ? – Sasha Jul 13 '11 at 15:56
The $A$ that I used is just any m*n matrix to elaborate the formula. – xenon Jul 13 '11 at 16:13

It sounds like you might be a bit confused about vectors. In your picture $e' = e$, since vectors (in this context) don't have an initial point. They are solely determined by their length and direction. Since they are equal, both $e$ and $e'$ either lie or don't lie in the nullspace.
What comes to $e$ belonging to the nullspace: Suppose that $A$ is a projection (i.e. $A^2 = A$), then if we let $e = b - Ab$, we have $$Ae = Ab - A^2 b = Ab - Ab = 0,$$ and thus $e$ is in the nullspace.
oh...the initial point isn't considered. So it is like the dotted orange line is the null space and this line is spanned across the whole stretch of vector $\vec{a}$ and while spanning, it isn't entirely a plane either? – xenon Jul 13 '11 at 16:17