= Solution
The <zero-composition subspaces of a linear map> are subspaces because the defining equations are homogeneous linear equations in $\beta$: they hold for the zero map and are preserved by addition and <scalar multiplication>. A map with $\beta\alpha=0$ vanishes on $\operatorname{im}\alpha$, so it factors uniquely through the <quotient vector space> $V/\operatorname{im}\alpha$. Conversely every map from that quotient to $U$ gives such a $\beta$. Therefore
$$
M^l(\alpha)\cong L(V/\operatorname{im}\alpha,U),\qquad
\boxed{\dim M^l(\alpha)=(\dim V-\operatorname{rank}\alpha)\dim U}.
$$
The condition $\alpha\beta=0$ says exactly that the <image> of $\beta$ lies in $\ker\alpha$. Consequently
$$
M^r(\alpha)\cong L(V,\ker\alpha),\qquad
\boxed{\dim M^r(\alpha)=\dim V\,\dim\ker\alpha}.
$$
Both dimension formulas use the fact that the space of maps between spaces of dimensions $p,q$ has dimension $pq$: <matrix> units in any chosen <bases> supply a <basis>.
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