Exercise 10.A.2

Proof. Suppose $A$ and $B$ are $n$-by-$n$ matrices. Let $V$ denote an $n$-dimensional vector space and choose a basis for $V$. Define $S,T\in L({𝔽}^n)$ such that $M(S)=A$ and $M(T)=B$, where this matrices are with respect to the chosen basis. We have

Therefore $\mathit{ST}=I$. Exercise 10 in section 3D shows that $\mathit{TS}=I$. Thus

The equation above completes the proof. □