# Homework Help: Similar Matrices to the k power

1. Mar 20, 2010

### Dustinsfl

If A and B are similar matrices, show that Ak and Bk are similar.
I am almost positive this has to be done by induction.

p(k):= Bk=S-1*Ak*S
p(k+1):= Bk+1=S-1*Ak+1*S
Assume p(k) is true.

I know I could take p(k) and multiply right by A but I don't think that will go any where.

2. Mar 20, 2010

### Tinyboss

If $$B=P^{-1}AP$$, then $$B^k=$$... ?

3. Mar 20, 2010

### Dustinsfl

I forgot to mention I am proving this for each positive integer or the natural numbers.

4. Mar 21, 2010

### VeeEight

You don't need induction here.
Answer the previous poster's question, what is Bk in terms of A?
If you can't get that right away, try it for a smaller integer. If B=P-1AP, what is B2? You should see how the answer follows.

5. Mar 21, 2010

### Dustinsfl

So I am going to obtain s inverse to the k a to the k and s to the k which is equal to b to the k

6. Mar 21, 2010

### Tinyboss

No, matrices don't commute under multiplication in general.

Edit: Although, if they did, you'd be home free, too...but they don't!

7. Mar 21, 2010

### Dustinsfl

What should I do then with (S-1*A*S)k then?

8. Mar 21, 2010

### Tinyboss

Write it out.

9. Mar 21, 2010

### VeeEight

Try small powers. For example, k=2
(XYZ)2=(XYZ)(XYZ)=XYZXYZ
Now try it for (P-1AP)2=.....

After you have that done you can generalize to the power of k.

10. Mar 21, 2010

### Dustinsfl

11. Mar 21, 2010

### VeeEight

Just multiply the term by itself.
(XYZ)2=(XYZ)(XYZ)=XYZXYZ

12. Mar 21, 2010

### Dustinsfl

Never mind my eyes were playing tricks on me I thought that was p- not to the power.

13. Mar 21, 2010

### Dustinsfl

I see how it works since the middle terms become I which goes away. I now that this will happen for where each last term meets the first time but I am still unsure how to prove it.

14. Mar 21, 2010

### VeeEight

If you see it for the smaller powers (for k=2,3.. etc), it is easy to generalize to the power of k. You can use induction if you want but it is not really needed.

15. Mar 21, 2010

### Tinyboss

You will use induction after all. But now that you see the pattern, you should be able to make it work.

16. Mar 21, 2010

### Dustinsfl

If I use my term to the k+1, what do I do after I separate it to term to the k times term to the first?

17. Mar 21, 2010

### VeeEight

(P-1AP)k+1 = (P-1AP)k(P-1AP). Then apply the induction hypothesis (assuming you proved a base case).

18. Mar 21, 2010

### Dustinsfl

This is the step I am stuck on.

19. Mar 21, 2010

### VeeEight

What exactly are you stuck on? Apply the induction hypothesis to the term (P-1AP)k. This will give you the familiar paring of the P and P-1 terms when you multiply it by P-1AP.

So you have (P-1AP)k(P-1AP) = (P-1AkP)(P-1AP) = P-1Ak(PP-1)AP = ......

20. Mar 21, 2010

### Staff: Mentor

By assumption, Bk = P-1AkP.

So Bk+1 = BkB = ?

21. Mar 21, 2010

### Dustinsfl

I was working the wrong side of the equation was the issue thanks.