- 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
2.16 Exercises
5. $\quad$ If $A$ is a square matrix, prove that $A^n A^m = A^{m+n}$ for all integers $m \geq 0,$ $n \geq 0.$
Proof. $\quad$ This proof will follow a similar approach to that used in Section 2.8, Exercise 21. That is, as a result of Theorem 2.5 (associative property of composition), if $A$ is a square matrix, we can express $A^nA^m$ as follows: \begin{align*} A^nA^m &= A^n(AA^{m-1}) \\ &= (A^nA)A^{m-1} \\ &= A(A(...(A)))A^{m-1} \\ &= A^{n+1}A^{m-1} \\ &\cdots \\ &=(A^{n + m - 1})A \\ &= A^{m + n} \quad \blacksquare \end{align*}