- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.8 Exercises
In Exercises 22 through 25, $S$ and $T$ denote functions with domain $V$ and values in $V.$ In general, $ST \neq TS.$ If $ST = TS,$ we say that $S$ and $T$ commute.
24.$\quad$ If $S$ and $T$ are invertible and commute, prove that their inverses also commute.
Solution. $\quad$ Let $S^{-1}$ and $T^{-1}$ denote the inverses of $S$ and $T,$ respectively. In Exercise 23, we showed that if $S$ and $T$ are invertible, then $ST$ is invertible and $(ST)^{-1} = T^{-1}S^{-1}.$ But if $S$ and $T$ commute, then $ST = TS$ and by extension, $(ST)^{-1} = (TS)^{-1}.$ But as we showed in the previous exercise, the inverse of $TS$ is the composition of the inverses, taken in reverse order. In other words, $(TS)^{-1} = S^{-1}T^{-1}.$ And since $(ST)^{-1} = (TS)^{-1},$ we have \begin{align*} T^{-1}S^{-1} &= S^{-1}T^{-1} \end{align*} Thus showing that if $S$ and $T$ are invertible and commute, then their inverses also commute. $\quad \blacksquare$