Exercise 8.B.9

Proof. Keep in mind that when we mention the size of an $n$-by-$n$ matrix here we mean $n$ and not $n$ times $n$.

Let $V$ be vector space whose dimension equals the size $A$ (or $B$, since they’re the same). Choose a basis of $V$ and define $S,T\in L(V)$ such that $M(S)=A$ and $M(T)=B$. Then $M(\mathit{ST})=\mathit{AB}$.

Let $d_j$ equal the size of $A_j$ (or $B_j$, because they’re the same). Consider the list consisting of the first $d_1$ vectors in the chosen basis. $A$ and $B$ show that the span of these vectors are invariant under $S$ and $T$. Similarly, the span of the next $d_2$ vectors after this list is also invariant under $T$. Continuing in this fashion, we see that there are $m$ distinct lists of consecutive vectors, with no intersections, in the chosen basis whose spans are invariant under $S$ and $T$.

Let $U_1,\dots ,U_m$ denote such spans. Clearly $M(S{\text{|}}_{U_j})=A_j$ and $M(T{\text{|}}_{U_j})=B_j$ for each $j$. Hence $M(S{\text{|}}_{U_j}T{\text{|}}_{U_j})=A_j{B}_j$ and so it easy to see that $M(\mathit{ST})$ (which equals $\mathit{AB}$) has the desired form. □