 Solution

Since $AB=0$, the image of a linear map represented by $B$ lies in the kernel of $A$. The matrix $B$ is nonzero, so this image of a linear map contains a nonzero vector; consequently $\dim \ker A\ge 1$. On the other hand, $A\ne 0$ gives $\mathrm{rank}A\ge 1$. The rank-nullity theorem in dimension two therefore givesThus the image of $A$ is a line through the origin, and $A$ maps onto that line.

Reversing the order of the matrix multiplication need not give zero. For example,gives $AB=0$ but $BA=B\ne 0$. This also shows why information about one composition of linear maps does not establish the corresponding property of the reversed composition.

Write $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Its rank is one, so its determinant is zero. Choose its adjugate matrix:Direct matrix multiplication gives $AC=CA=(ad-bc)I=0$. Since the entries of $C$ are the entries of the nonzero matrix $A$, rearranged with possible sign changes, $C\ne 0$. This establishes both required annihilation properties.