Question

If $AB=BA$ for any two square matrices, prove by mathematical induction that $(AB{)}^n=A^n{B}^n$

Answer 1

Checking that statement is true for $n=1$.
Let $P\left(n\right):(AB{)}^n=A^n{B}^n$
For $n=1$
$P\left(1\right):(AB{)}^1=A^1{B}^1$
$\Rightarrow AB=AB$
$\therefore P\left(1\right)\text{is true.}$

Assume $P\left(n\right)$ be true for $n=k$.
(Where $k\in N)$
So, $P\left(k\right):(AB{)}^k=A^k{B}^k,k\in N$ ... (1)

Proving for $n=k+1$
i.e. to prove $P(k+1):(AB{)}^{k+1}=A^{k+1}{B}^{k+1}$
$(AB{)}^{k+1}=(AB{)}^k\left(AB\right)$
$\Rightarrow (AB{)}^{k+1}=A^k{B}^k(AB)$
$\Rightarrow (AB{)}^{k+1}=A^k{B}^{k-1}B(AB)$
$\Rightarrow (AB{)}^{k+1}=A^k{B}^{k-1}ABB[\because AB=BA]$
$\Rightarrow (AB{)}^{k+1}=A^k{B}^{k-1}A{B}^2$

Repeating same step, we get
$\Rightarrow (AB{)}^{k+1}=A^k{B}^{k-1}A{B}^3$
Following this several times , we get
$\Rightarrow (AB{)}^{k+1}=A^{k+1}{B}^{k+1}$
$\therefore P(k+1)\text{is true}$.

So $P\left(n\right)$ is true for all $n\in N$.
Hence $(AB{)}^n=A^n{B}^n$, for all $n\in N$