Prove that any vector can be written as the sum of any three non-coplanar vectors I'm trying to prove the above in the following form:
$ \boldsymbol{V} = (\boldsymbol{V} \centerdot \boldsymbol{a}^1)\boldsymbol{a}_1 + (\boldsymbol{V} \centerdot \boldsymbol{a}^2)\boldsymbol{a}_2 + (\boldsymbol{V} \centerdot \boldsymbol{a}^3)\boldsymbol{a}_3 $,
where $ \boldsymbol{a}_1 $, $ \boldsymbol{a}_2 $ and $ \boldsymbol{a}_3 $ are any three non-coplanar vectors, and $ \boldsymbol{a}^1 $, $ \boldsymbol{a}^2 $ and $ \boldsymbol{a}^3 $ are corresponding "reciprocal" vectors, such that:
$ \boldsymbol{a}^1 = \dfrac{\boldsymbol{a}_2 \times \boldsymbol{a}_3}{\boldsymbol{a}_1 \centerdot \boldsymbol{a}_2 \times \boldsymbol{a}_3} $,
$ \boldsymbol{a}^2 = \dfrac{\boldsymbol{a}_3 \times \boldsymbol{a}_1}{\boldsymbol{a}_1 \centerdot \boldsymbol{a}_2 \times \boldsymbol{a}_3} $,
$ \boldsymbol{a}^3 = \dfrac{\boldsymbol{a}_1 \times \boldsymbol{a}_2}{\boldsymbol{a}_1 \centerdot \boldsymbol{a}_2 \times \boldsymbol{a}_3} $.
Is there a way to complete the proof without doing a lengthy expansion? What is an intuitive reason for why the scaling factors $ \boldsymbol{V} \centerdot \boldsymbol{a}^i $ work?
For example, taking $ \boldsymbol{V} \centerdot \boldsymbol{a}^1 $ I've got as far as supposing that if $ \boldsymbol{a}_2 \times \boldsymbol{a}_3 $ is a vector representing the base of parallelepiped $ \boldsymbol{a}_1 \centerdot \boldsymbol{a}_2 \times \boldsymbol{a}_3 $, then dividing top and bottom, $ \boldsymbol{V} \centerdot \boldsymbol{a}_2 \times \boldsymbol{a}_3 $ and $ \boldsymbol{a}_1 \centerdot \boldsymbol{a}_2 \times \boldsymbol{a}_3 $, by $ |\boldsymbol{a}_2 \times \boldsymbol{a}_3| $, leaves the length of $ \boldsymbol{V} $ relative to the length of $ \boldsymbol{a}_1 $ as measured along the vector $ \boldsymbol{a}_2 \times \boldsymbol{a}_3 $. I'm not sure though how to relate this to the rectangular coordinate system common to $ \boldsymbol{V} $ and $ \boldsymbol{a}_1 $.
 A: What you're looking for is simply a proof that a 3-dimensional vector space can be expressed as the span of three linearly independent vectors.
I suppose you know that any 3-vector can be expressed as a weighted sum of the canonical basis vectors, $\hat{x}$, $\hat{y}$, and $\hat{z}$. You can see that this is the case because you need three components to specify a 3D-vector, and each of those vectors corresponds to one of the components.
What you perhaps don't know is that the basis vectors you use to correspond to the three needed coordinates don't need to be orthogonal, just linearly independent: as long as none of them can be expressed in terms of the others, they are sufficient to span all of the vector space.
You can find proofs of this in plenty of places; check any introduction to linear algebra. Your definitions of $\mathbb{a}^i$ are irrelevant to the discussion, except to verify that the three vectors in question are in fact non-coplanar.
By the way, you may be aware of the fact that invertible matrices are isomorphims on a vector space, i.e. multiplying all of $\mathbb{R}^3$ by some 3x3 matrix $M$ gives you back a copy of $\mathbb{R}^3$ again (though the coordinates of everything are different). If you then multiply by $M^{-1}$, you get back to the same coordinatization as where you started, in the same way that $a^{-1} b a = b$ again in scalar arithmetic. Now it just so happens that there is always a matrix $M$ which will send your three vectors $\mathbf{a}^i$ to the canonical unit vectors $\hat{x}, \hat{y}, \hat{z}$; it is the inverse of the matrix with $\mathbf{a}^i$ as the $i^\text{th}$ column. Therefore you may intuitively prove the required fact by sending your vectors to the usual unit vectors, seeing that clearly those three are needed to express any 3-vector, and then using $M^{-1}$ to send the coordinates back to where you started. (This isn't really a proof, just some intuition. As I mentioned, the proof you're looking for is published anywhere that teaches linear algebra.)
A: If you rewrite the starting identity as
$$\boldsymbol{V} = b_1\boldsymbol{a}_1 + b_2\boldsymbol{a}_2 + b_3\boldsymbol{a}_3 $$
and put it into matricial form
$$\boldsymbol{V} = \boldsymbol{A}\, \boldsymbol b $$
where $\boldsymbol{A}$ is the matrix composed by the vertical vectors $\boldsymbol{a}_k$, and $\boldsymbol b $ is the vertical vector with components $b_k$, then 
$$det(\boldsymbol {A})={\boldsymbol{a}_1 \centerdot \boldsymbol{a}_2 \times \boldsymbol{a}_3}$$
and your starting identity is just an alternative way to write the Cramer's rule, in solving $\boldsymbol{V} = \boldsymbol{A}\, \boldsymbol b $ for $\boldsymbol b $.
