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I can't see how to do this. I cant see how those two vectors can span a subspace of $\mathbb{R}^3$ when they both have 4 components.

Here's what I've done. I tried to find the column space of these two vectors by putting them into a $4\times2$ matrix

$$ \left[\begin{array}{rr|r} -4 & 2 & a \\ 2 & 2 & b \\ 2 & 2 & c \\ -4 & 0 & d \end{array}\right] $$

I then row-reduced and got the following results.

$$ \left[\begin{array}{rr|l} -4 & 2 & a \\ 0 & 6 & 2b - a\\ 0 & 0 & c - b \\ 0 & 0 & 3d - 2a + 2b \end{array}\right] $$

So then I used the equation

$3d - 2a + 2b = 0$ to obtain two $1\times3$ vectors that span $\mathbb{R}^3$

$a = b + \frac{3}{2}d$

$\begin{bmatrix} a \\ b \\ d \end{bmatrix} = \operatorname{span} \left\{\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} \frac{3}{2} \\ 0 \\ 1 \end{bmatrix}\right\}$

But I cant project a $4\times1$ onto these vectors because I cant dot a $1\times4$ vector with a $1\times3$ vector.

I have a feeling I have got the workings of this question all wrong. What am I supposed to be doing?

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It's a typo, they meant $\Bbb R^4$. – David Mitra Apr 11 '12 at 14:41
Stupid typo. What is the correct procedure now, is it getting the projection of v against of each of those vectors and then adding the results together? – Jim_CS Apr 11 '12 at 15:42
I projected v onto both of those vectors, added the results and got $(14/10, 8/10, 8/10, 4/10)$ but it is showing up as an incorrect answer... – Jim_CS Apr 11 '12 at 15:49
You need to first use Gram-Schmidt (or something similar) to find a basis for the two-dimensional sub-space that is orthogonal. Then you can project onto each of the basis directions and add. – Willie Wong Apr 11 '12 at 15:58
Could somebody explain to me what is actually being asked in this question? I think I don't understand what is being asked... – Jim_CS Apr 11 '12 at 17:10

Here's another approach. Let $v_1, v_2$ be the given basis for the subspace $V$. Let $p$ be the desired projection. Since $p \in V$, it can be written as $p=[v_1 v_2] x$, where $x \in \mathbb R^2$. The problem then reduces to finding $x$.

Since $p$ is the projection of $v$ on $V$, then $p-v$ must be perpendicular to $V$, or in other words: $<v_i,p-v>=0$, for $i=1,2$. Substituting for $p$ and rewriting slightly gives: $ v_1^T[v_1 v_2] x = v_1^T v$, $ v_2^T[v_1 v_2] x = v_2^T v$.

This reduces to two simple equations $40 x_1 = -20$, $12 x_2 = 6$, and so we have $x_1=-\frac{1}{2}, x_2 = \frac{1}{2}$. Using $p=[v_1 v_2] x$, we get $p^T = (3,0,0,2)$.

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