Q:

Linear Algebra Ordered Basis

If $V$ is a subspace of $\Bbb{R}^3$ with ordered basis $B = (v_1,v_2)$, and let $v_3,v_4 \in V$. $A$ is the $3 \times 4$ matrix whose $i$th column is $v_i$. Find a $3 \times 2$ matrix $B$ and a $2 \times 4$ matrix $C$ such that $A = BC$.

We are talking about orthogonality in class and I think it has something to do with that.  I understand matrix multiplication but have now idea where to start with this other than $A=BC$. Any help would be appreciated! 

A:

Let
$$T : \Bbb{R}^4 \to V : (a, b, c, d) \mapsto av_1 + bv_2 + cv_3 + dv_4.$$
Note $T$ is linear.  Further, let
$$J : V \to \Bbb{R}^3$$
be the inclusion map from $V$ into $\Bbb{R}^3$. That is, $J(x) = x$ (like the identity map) for all $x \in V$, but with the codomain being a larger space than $V$. Naturally, $J$ is linear too. 
Let $E$ be the standard basis for $\Bbb{R}^4$, $F$ be the standard basis for $\Bbb{R}^3$, and $B = (v_1, v_2)$, the basis given for $V$. We have
$$[JT]_{F \leftarrow E} = [J]_{F \leftarrow B} [T]_{B \leftarrow E}.$$
Note also that $[JT]_{F \leftarrow E} = A$, as transforming each standard basis vector by $T$ yields $v_i$, which $J$ includes in $\Bbb{R}^3$, and hence the $i$th column of $[JT]_{F \leftarrow E}$ is $v_i$. Thus, we can simply take $B = [J]_{F \leftarrow B}$ and $C = [T]_{B \leftarrow E}$.

