projecting a point onto a vector Problem: Define $P$ to be the projection onto the vector $($2$,$-3$)$. Find the matrix that represents $P$.
I'm a new student to linear algebra and I've seen projections onto lines, planes, surfaces, etc. But I've never seen projections onto a vector. Is that what this problem is asking? When the problem says that $P$ is the projection onto a vector, why doesn't it specify what exactly is being projected onto it?
Some clarification would be very helpful.
 A: The projection is found, by asking how much does the pointer $x$ point in the direction of $v$, and than heading in the direction $v$ by that amount. 
$$proj_v(x)=v\cdot x \frac{v}{|v|}$$
which is 
$$proj_v(x_i)=\sum\limits_{j=0}^Jv^jx_j\frac{v_i}{|v|}=\sum\limits_{j=0}^J\frac{1}{|v|}(v^jv_i)x_j$$
By inspecting the above expression, we see that 
$$\frac{1}{|v|}(v^jv_i)$$
can be seen as a matrix with indices $j$ for rows  and $i$ for columns. Thus
$$P^j_i=\frac{1}{|v|}(v^jv_i)$$
which means
$$proj_v(x)=P\cdot x$$
A: Expanding on my hint in the comment above: what would the projection $Q$ onto $(1,0)$ look like? Well, if $v = (a,b) = a(1,0) + b(0,1)$ then the projection of $v$ is
$$Qv = aQ(1,0) + bQ(0,1) = a(1,0) + b(0,0) = a(1,0)$$
by linearity of $Q$, the fact that $Q$ leaves $(1,0)$ invariant but sends $(0,1)$ to the zero vector.
Now for your problem, let's write $v = (a,b) = \alpha(2,-3) + \beta(3,2)$ for some appropriate choice of $\alpha, \beta$. Notice that $P(2,-3) = (2,-3)$ but $P(3,2) = (0,0)$. Hence
$$Pv = \alpha(2,-3)$$
So find how to write $\alpha, \beta$ in terms of $a, b$ and you're most of the way there.

To give a specific example, suppose $v = (1,1)$. Then
$$v = - \frac{1}{13}(2,-3) + \frac{5}{13}(3,2)$$
Thus $Pv = - \frac{1}{13}(2,-3)$

ADDED (see below):
Solving for $\alpha, \beta$ using your favorite method, we find that 
$$\alpha = \frac{2a - 3b}{13}, \ \beta = \frac{3a + 2b}{13}$$
Thus as 
$$P\left(\begin{matrix} a \\ b \end{matrix}\right) = \alpha\left(\begin{matrix} 2 \\ -3 \end{matrix}\right) =\frac{2a - 3b}{13} \left(\begin{matrix} 2 \\ -3 \end{matrix}\right) = \left(\begin{matrix} \frac{4}{13}a -\frac{6}{13}b \\ -\frac{6}{13}a + \frac{9}{13}b \end{matrix}\right) $$
We can now read off the entries of $P$,
$$P = \left(\begin{matrix} \frac{4}{13} & -\frac{6}{13} \\ -\frac{6}{13} & \frac{9}{13} \end{matrix}\right)$$
