1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Similar Matrices to the k power

  1. Mar 20, 2010 #1
    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. jcsd
  3. Mar 20, 2010 #2
    If [tex]B=P^{-1}AP[/tex], then [tex]B^k=[/tex]... ?
  4. Mar 20, 2010 #3
    I forgot to mention I am proving this for each positive integer or the natural numbers.
  5. Mar 21, 2010 #4
    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.
  6. Mar 21, 2010 #5
    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
  7. Mar 21, 2010 #6
    No, matrices don't commute under multiplication in general.

    Edit: Although, if they did, you'd be home free, too...but they don't!
  8. Mar 21, 2010 #7
    What should I do then with (S-1*A*S)k then?
  9. Mar 21, 2010 #8
    Write it out.
  10. Mar 21, 2010 #9
    Try small powers. For example, k=2
    Now try it for (P-1AP)2=.....

    After you have that done you can generalize to the power of k.
  11. Mar 21, 2010 #10
    I not sure about this part.
  12. Mar 21, 2010 #11
    Just multiply the term by itself.
  13. Mar 21, 2010 #12
    Never mind my eyes were playing tricks on me I thought that was p- not to the power.
  14. Mar 21, 2010 #13
    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.
  15. Mar 21, 2010 #14
    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.
  16. Mar 21, 2010 #15
    You will use induction after all. But now that you see the pattern, you should be able to make it work.
  17. Mar 21, 2010 #16
    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?
  18. Mar 21, 2010 #17
    (P-1AP)k+1 = (P-1AP)k(P-1AP). Then apply the induction hypothesis (assuming you proved a base case).
  19. Mar 21, 2010 #18
    This is the step I am stuck on.
  20. Mar 21, 2010 #19
    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 = ......
    Your result follows from this.
  21. Mar 21, 2010 #20


    Staff: Mentor

    By assumption, Bk = P-1AkP.

    So Bk+1 = BkB = ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook