Hi Ross.
Try rightmultiplying 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
AB^{n} = B^{n} 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 AB^{2} = B^{2} A. You now know that AB^{2} = B^{2} A so look at Sue's argument again and in the first step replace AB = BA by AB^{2} = B^{2} A.
You get
Try rightmultiplying both sides by B:
(AB^{2})B = (B^{2}A)B
but matrix multiplication is associative, so
(B^{2}A)(B) = (B^{2})(AB)
and continuing her argument you end with AB^{3} = B^{3} A.
Now that you know that AB^{3} = B^{3} A you can replace Sue's first step by
Try rightmultiplying both sides by B:
(AB^{3})B = (B^{3}A)B
and continue her argument to show that AB^{4} = B^{4} 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 AB^{n} = B^{n} 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 AB^{n} = B^{n} A is true for n = k, that is AB^{k} = B^{k}A. I want then to show it is true for n = k + 1. That is I want to show that AB^{k+1} = B^{k+1}A.
Try rightmultiplying both sides by B:
(AB^{k})B = (B^{k}A)B
Continue with Sue's argument to show that AB^{k+1} = B^{k+1}A.
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
