Subject: Induction
Name: Ross
Who are you: Parent

Hi I am a parent of a 18 year old who is having problems with Discrete Mathematics and I'm not sure how to do it myself. So if you could help help me out I would greatly appreciate.

Question:
Suppose that A and B are square matrices with the property AB= BA. Show that ABⁿ = Bⁿ A for every positive integer n.

Thanks

Hi Ross.

Try right-multiplying both sides by B:

(AB)B = (BA)B

but matrix multiplication is associative, so

(BA)(B) = (B)(AB)

and from the given equality, AB = BA, that means

(B)(AB) = (B)(BA).

So we've shown (AB)(B) = (B)(BA).

Can you take it from here?
Stephen La Rocque>

Ross wrote back.

Hi,

I understand everything you solved but I still don't know how to "Show that
ABⁿ = Bⁿ A for every positive integer n." This is the part that is confusing me. So if you could let me know and possibly take it a little farther for me I would greatly appreciate.

Thanks,
Ross

Ross,

The technique for proving this fact is called Mathematical Induction. There are two steps in the proof. The first is to verify that the statement is true when n = 1. In your problem this is straightforward since you are told that for the matrices A and B,
AB = BA.

What Sue did was to show you how to use this fact and the fact that matrix multiplication is associative to show that AB² = B² A. You now know that AB² = B² A so look at Sue's argument again and in the first step replace AB = BA by AB² = B² A.

Try right-multiplying both sides by B:

(AB²)B = (B²A)B

but matrix multiplication is associative, so

(B²A)(B) = (B²)(AB)

and continuing her argument you end with AB³ = B³ A.

Now that you know that AB³ = B³ A you can replace Sue's first step by

Try right-multiplying both sides by B:

(AB³)B = (B³A)B

and continue her argument to show that AB⁴ = B⁴ A.

Now that you know that ...

Mathematical Induction gives you a way to replace this infinite chain of steps by two steps. Step one is completed, that is when n = 1. For the second step, called the inductive step, you assume that you know the statement is true for some value of n. By this I mean that ABⁿ = Bⁿ A is true for same value on n. I am going to assume it is true for the value k. Here is my inductive step.

Suppose that ABⁿ = Bⁿ A is true for n = k, that is AB^k = B^kA. I want then to show it is true for n = k + 1. That is I want to show that AB^k+1 = B^k+1A.

Try right-multiplying both sides by B:

(AB^k)B = (B^kA)B

Continue with Sue's argument to show that AB^k+1 = B^k+1A.

The proof is then complete. You have shown that the statement is true for n = 1 and then, using the inductive step it is true for n = 1 + 1 = 2. Once you know it is true for
n = 2 the inductive step shows it is true for n = 2 + 1 = 3 and so on.

I hope this helps,
Penny