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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
    (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.
     
  11. Mar 21, 2010 #10
    I not sure about this part.
     
  12. Mar 21, 2010 #11
    Just multiply the term by itself.
    (XYZ)2=(XYZ)(XYZ)=XYZXYZ
     
  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

    Mark44

    Staff: Mentor

    By assumption, Bk = P-1AkP.

    So Bk+1 = BkB = ?
     
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