Question

If A and B are square matrices of the same order such that AB=BA, then prove by induction that A$B^n$=$B^n$A. Further, prove that $(AB{)}^n$ =$A^n$$B^n$ for all $n\in N$.

Answer 1

(a)Given, that AB=BA...........(i)
We want to prove that $A{B}^n=B^{n}A.........\left(ii\right)$
For n=1, Eq. (ii)is obviously true.
Let Eq. (ii)be true for a positive integer n=m.
i.e. $A{B}^m=B^{m}A..........\left(iii\right)$
Then, for n=m +1, $A{B}^{m+1}=A\left(B^{m}B\right)=\left(A{B}^m\right)B$
(Associative law of matrix multiplication)
$=\left(B^{m}A\right)B$ [Using Eq. (iii)]
$=B^m\left(AB\right)=B^m\left(BA\right)$ [Using Eq.(i)]
$=\left(B^{m}B\right)A=B^{m+1}A$. Hence, by induction Eq. (ii)is true for all n \in N.

(b)Here, given that AB=BA ....(i)
We want to prove that $(AB{)}^n=A^n{B}^n.......(ii)$
For n =1, Eq. (ii)is obviously true. [\because from Eq. (i)]
Let Eq. (ii) be true for a positive integer n =m i.e., $(AB{)}^m=A^m{B}^m......(iii)$
Then, for $n=m+1,(AB{)}^{m+1}=(AB{)}^m\left(AB\right)=\left(A^m{B}^m\right)\left(AB\right)$
[Using Eq. (iii)]
$=A^m\left(B^{m}A\right)B=A^m\left(A{B}^m\right)B.$
$(\therefore A{B}^n=B^{n}A$ for all $n\in N$ whenever AB=BA)
$=\left(A^{m}A\right)\left(B^{m}B\right)=A^{m+1}{B}^{m+1}$
Hence, by induction Eq. (ii)is true for all $n\in N.$