The following question came up on a recent exam that I took.
Let $U$, $V$ and $W$ be complex vector spaces such that $\dim V = 4$. Let $A$ be a linear transformation from $U$ to $V$ such that the dimension of the kernel of $A$ is 2. Let $B$ be a linear transformation from $V$ to $W$, and consider the transformation $BA$ from $U$ to $W$ such that the dimension of the kernel $BA$ is 3. List all possibilities of the dimension of the range (rank) of $BA$.
Proposed Proof:
We consider multiple cases by considering the dimension of the range of the transformation $A$. If rank$(A)$ = 0, then $A$ is the zero map. In this case, for every $u \in U$ such that, we have, \begin{equation} BA(u) = B(A(u)) = B(0_V) = 0_W. \end{equation} So, $BA$ is the zero map, and the rank is 0.
If rank$(A)$ is 1, we have that $BA$ is the zero map is $B$ is the zero map. In this case, rank$(BA) = 0$. Let $B$ not be a zero map. We claim that $BA$ must still be the zero map. Assume $BA$ isn't the zero map. Let $u \in R(A)$ such that $u \neq 0$, and $B(u) \neq 0$. Since $R(A)$ is a 1-dimensional subspace, we have that, any $v \in R(A)$, $v \neq 0$, is such that, $v = \alpha u, \alpha \neq 0$. Then, $B(v) = \alpha B(u) \neq 0$. Hence, the kernel of $BA$ remains of dimension 2 because all vectors outside of the kernel of $A$, which are mapped into $R(A)$, are mapped to non-zero vectors, a contradiction. Hence, $BA$ must be the zero map.
Now let rank$(A) = 2$. WLOG, let $B$ not be the zero map. Let $\beta = \{ v_1, v_2 \}$ be a basis for $R(A)$. Then, \begin{equation} B(R(A)) = span (R(A)) = \alpha_1 B(v_1) + \alpha_2 B(v_2), \end{equation} where $B(v_1) \neq 0$ or $B(v_2) \neq 0$. If both of them are non-zero, then, as before, the dimension of the kernal would not increase from 2 to 3. So, one of them is zero. If exactly one is zero, we will get rank$(BA) = 1$. If $B$ is the zero map the rank is $0$.
If rank$(A) = 3$, by arguments as above, a 1-dimensional subspace of $R(A)$ must be mapped to $0_W$. So, rank$(BA)$ is either $0, 1, 2$. If rank$(A)$ is 3, then rank$(BA)$ is either $0,1,2,3$.
I think the argument is correct, but, due to time constraints, I wrote a not so pristine proof. I think a lot of these cases could have been handled directly by extending basis of kernel of $B$ to kernel of $BA$, and referring to the rank-nullity theorem implies these are the only possibilities for rank$(BA)$. I have tried reproducing the proof by including only those details I wrote on the paper. There were some other details I wrote which I didn't end up using to complete the argument.
This quesion was for 10 points, so I am trying to see if I have at least some parts correct.