- 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
21.$\quad$ Let $T: V \rightarrow V$ be a function which maps $V$ into itself. Powers are defined inductively by the formulas $T^0 = I,$ $T^n = TT^{n-1}$ for $n \geq 1.$ Prove that the associative law for composition implies the law of exponents: $T^m T^n = T^{m+n}.$ If $T$ is invertible, prove that $T^n$ is also invertible and that $(T^n)^{-1} = (T^{-1})^n.$
Proof. $\quad$ By the associative law of composition, we can rewrite $T^mT^n$ as follows: \begin{align*} T^mT^n &= T^m(TT^{n-1}) \\ &= (T^mT)T^{n-1} \\ &= T(T(...(T)))T^{n-1} \\ &= T^{m+1}T^{n-1} \\ &\cdots \\ &= T^{m + n - 1}T \\ &= T^{m + n} \end{align*} If $T$ is invertible, we have $T^{-1}T = I.$ Expanding $T^n$ and applying $T^{-1}$ we get \begin{align*} T^{-1}(TT^{n-1}) &= (T^{-1}T)T^{n-1} \\ &= IT^{n-1} \\ &=T^{n-1} \end{align*} Expanding the right-hand side and applying the inverse $n-1$ additional times, we find that $(T^n)^{-1} = (T^{-1})^n. \quad \blacksquare$