Standard Matrix of an Oblique Projection through a point Q : Let P: R2 --> R2 be the linear transformation which projects points onto the line 2x-y=0, not orthogonally, but in the direction of :
V = $\begin{bmatrix}7\\4\end{bmatrix}$
Find the standard Matrix.
Now I've yet to see an example of non-orthogonal projections in R2 in a specific direction, however I have -and know - how to compute projections orthogonally through a plane by using the outer product of the normal. Can anyone explain how this relates, or rather what differs in this case?
 A: Let's first consider the orthogonal projection case and find its matrix $P$.
We have that
$$
A = \begin{bmatrix}1\\2\end{bmatrix}
$$
is a basis for the range of the projection and then have
$$
P = A(A^\top A)^{-1}A^\top = \frac15 \begin{bmatrix}1\\2\end{bmatrix} \begin{bmatrix}1 & 2\end{bmatrix} = \begin{bmatrix}1/5 & 2/5\\2/5 & 4/5\end{bmatrix}.
$$
Now let's find the oblique projection $Q$ through $V=[7\;4]^\top$.
Now let
$$
B = \begin{bmatrix}-4\\7\end{bmatrix}
$$
be a basis for the orthogonal complement of the null space of the projection
We then have
$$
Q = A (B^\top A)^{-1} B^\top = \frac1{10}\begin{bmatrix}1\\2\end{bmatrix} \begin{bmatrix}-4 & 7\end{bmatrix} = \frac{1}{10} \begin{bmatrix}-4 & 7\\-8 & 14\end{bmatrix}.
$$
Then we can see that, for example, $Q$ projects the point $\begin{bmatrix}7\\-10\end{bmatrix}$ to $\begin{bmatrix}-9.8\\-19.6\end{bmatrix}$.
Visually, what's happening is that we're looking for the intersection between the line $y=2x$ and the line (in point-slope form) starting at $(7,-10)$ with slope $4/7$, i.e., $y-(-10) = \tfrac47(x-7)$.
Seen another way, we're looking to solve the system
\begin{align}
\begin{bmatrix} 7 \\ -10 \end{bmatrix} + c\begin{bmatrix} 7 \\ 4 \end{bmatrix} = d \begin{bmatrix} 1 \\ 2 \end{bmatrix}
\end{align}
where $c$ and $d$ are constants that tell us "how far to go" for the lines to intersect.
We use the specific form of $B$ that is the orthogonal complement of $\begin{bmatrix} 7 \\ 4 \end{bmatrix}$ since this normal vector defines the hyperplane (or line) $\begin{bmatrix} 7 \\ 4 \end{bmatrix}$ whose direction we want to project along.
