Describe the image and kernel of the projection function & verify the rank-nullity theorem:
The projection function is defined as $P_B(M) = {(M \cdot B)}\ /\ B$ where M is a multivector in $GA^n$ and B is a k-blade.
My work so far: So the image is just $\mathrm {Im}(P_B) = \{{(M \cdot B)}\ /\ B\ |\ M \mathrm {\ is\ a\ multivector\ in\ GA^n\ and}\ B \mathrm {\ is\ a\ blade}\}$. $\ $I can't figure out any way to simplify that. The kernel is $K(P_B) = \{M\ | \ {(M \cdot B)}\ /\ B = 0\} = \{M\ |\ (M \cdot B) = 0\}$. $\ $ I don't know how else to describe these.
Then the rank-nullity theorem is about their dimensions. The dimension of the kernel would be at most n - grade($B$) because all $M \cdot B \neq 0$ are in the subspace of $R^n$ represented by $B$ which is grade($B$)-dimensional. (I'm using left contraction for the dot product.)
I'm not sure how to figure out exactly what the dimension of the kernel or image is though.
Macdonald doesn't go far enough in describing the kernel and image, I think. It's known from vanilla linear algebra that the kernel and image are subspaces; using geometric algebra, we can say that the kernel and image are blades, since we can use blades to represent subspaces.
So what we really want to do is take the pseudoscalar $i$ ($n$-blade) and factor it into two blades $i = KI$, where $K$ is the kernel, and $P(I)$ is the image.
It's hard to do this systematically; most of the literature for finding the kernel of a linear map focuses on row reduction of a matrix, a specifically coordinate-based solution. I think Macdonald intended you to look at this problem geometrically: this is a projection onto $B$, so the kernel is naturally the orthogonal complement of $B$, that is, $iB$. Similarly, the projection forces all vectors into $B$ (or to zero), so the image is $B$.
But, this is one of those problems where the solution is so obvious on the one hand that I'm having a hard time figuring out a good and valid proof for it. I suspect one would start from the top down: first, see if $P(i)$ is nonzero. If it is nonzero, then there is no kernel, and the image is $i$. If it is zero, then look at all combinations of $n-1$-blades. If there is a nonzero combination, then there is a common vector underlying all combinations that do yield zero.
You'll notice that the factorization $i = KI$ implicitly gives us the rank-nullity theorem, as the grades of $K$ and $I$ must add to the grade of $i$ (which is $n$), and linear maps are grade-preserving.