- 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.
22.$\quad$ If $S$ and $T$ commute, prove that $(ST)^n = S^n T^n$ for all integers $n \geq 0.$
Solution. $\quad$ Suppose powers of $S$ and $T$ are defined as in Exercise 21. If $n = 0,$ $(ST)^n = I$ and the proof holds trivially. If $n = 1$ we can see that $(ST)^n = ST = S^nT^n.$ Now, assume this is true for some integer $k \geq 1,$ we will now prove that it is also true for $k + 1.$ When $n = k,$ we have \begin{align*} (ST)^k &= S^kT^k \end{align*} Then, by the inductive definition of the powers of $S$ and $T,$ we have \begin{align*} (ST)^{k+1} &= ST(ST)^k \\ &=(ST)S^kT^k \end{align*} But because $S$ and $T$ commute, we have \begin{align*} (ST)S^kT^k &= S(TS)S^{k-1}T^k \\ &= S(ST)S^{k-1}T^k \\ &= S^2TS^{k-1}T^k \\ &= S^2(TS)S^{k-2}T^k \\ &\cdots \\ &= S(S^kT^{k+1}) \\ &= S^{k+1}T^{k+1} \end{align*} Thus, by induction, we have proven this for all integers $n \geq 0. \quad \blacksquare$