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. You get 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 ABk = BkA. I want then to show it is true for n = k + 1. That is I want to show that ABk+1 = Bk+1A. Try right-multiplying both sides by B: (ABk)B = (BkA)B Continue with Sue's argument to show that ABk+1 = Bk+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