Question

If A and B are square matrices of the same order such that AB = BA , then prove by induction that . Further, prove that for all n ∈ N

Answer 1

Use the mathematical induction to prove the result AB n = B n A.

Consider the mathematical induction for n positive integer as P( n ) if,

P( n )= AB n ︸ LHS = B n A ︸ RHS (1)

Put n=1 to the left hand side of equation (1).

L.H.S.= AB 1 =AB

Put n=1 to the right side of equation (1).

R.H.S.= B 1 A =BA

Thus, left hand side is equal to the right hand side. So, the result P( n ) is true for n=1.

Assume the result P( k ) as true.

AB k ︸ LHS = B k A ︸ RHS (2)

Replace k by k+1 in equation (2) to prove the result P( k+1 ) is true.

P( k+1 ): if AB=BA

Then,

AB k+1 ︸ LHS = B k+1 A ︸ RHS

Consider left hand side,

AB k+1 =A( B k B ) =( AB k )B =( B k A )B = B k ( AB )

Given that AB=BA, then

AB k+1 = B k ( BA ) =( B k B )A = B k+1 A =R.H.S.

Therefore, the result P( k+1 ) is true when P( k ) is true.

By the mathematical induction prove that P( n ) is true for all n∈N.

Hence, if AB=BA, then AB n = B n A, where n∈N.

Now, use the mathematical induction to prove the result ( AB ) n = B n A n .

Consider the mathematical induction for n positive integer as P( n ) if,

P( n )= ( AB ) n ︸ LHS = B n A n ︸ RHS (1)

Put n=1 to the left hand side of equation (1).

L.H.S.= ( AB ) 1 =AB

Put n=1 to the right side of equation (1).

R.H.S.= B 1 A 1 =BA

Thus, left hand side is equal to the right hand side. So, the result P( n ) is true for n=1.

Assume the result P( k ) as true.

( AB ) k ︸ LHS = B k A k ︸ RHS (2)

Replace k by k+1 in the equation (2) to prove the result P( k+1 ) is true.

P( k+1 ): if AB=BA

Then,

( AB ) k+1 ︸ LHS = B k+1 A k+1 ︸ RHS

Consider left hand side,

( AB ) k+1 = ( AB ) k AB =( A k B k )( AB ) = A k B k ( BA )      [ AB=BA ] = A k ( B k B )A

Simplify the above expression,

( AB ) k+1 = A k ( B k B )A = A k ( B k+1 A ) = A k ( AB k+1 )      [ AB n = B n A ] =( A k A ) B k+1

Further simplify the above expression,

( AB ) k+1 = A k+1 B k+1 =R.H.S.

Therefore, the result P( k+1 ) is true when P( k ) is true.

By the mathematical induction prove that P( n ) is true for all n∈N.

Hence, if AB=BA, then ( AB ) n = B n A n , where n∈N.