I'm currently trying to prove Cramers rule with the wedge product definition of a determinant. The proof hinges on the fact that if M is some map, then we can write:
$$ M\left(\vec a \wedge \vec b\right) = M\left( \vec a \right) \wedge M\left( \vec b \right) $$
However, I'm at a loss here, because I really don't understand why this should be true. IF M is a linear map, I thought linearity implies that $M(a+b) = M(a)+M(b)$, can anybody explain why the above identity is true?
Thanks in advanced
This is perhaps a slight abuse of notation in that, as John pointed out, $M$ acting on a bivector $a \wedge b$ seems incompatible with the notion that $M$ acts on ordinary vectors.
A possible resolution to the notation issue would be to write, for vectors $v_1, v_2, \ldots, v_k$,
$$M_{(k)} (v_1 \wedge v_2 \wedge \ldots \wedge v_k) \equiv M(v_1) \wedge M(v_2) \wedge \ldots \wedge M(v_k)$$
And hence, given a linear operator $M$, we also define several other linear maps $M_{(2)}, M_{(3)}$, and so on. Nevertheless, it's pretty common to just say $M(a \wedge b) \equiv M(a) \wedge M(b)$ since there's little chance of getting confused about what map you're applying.
Since each of the $M_{(k)}$ is technically distinct from the others, I think you should be able to accept that this is merely a convenient definition? That we choose this definition might be motivated from the tensor product, but I must stress it is merely a convenient choice until one asks why and how this definition gives us $M_{(n)}(\epsilon) = (\det M) \epsilon$ for any $n$-vector $\epsilon$.