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Similar Matrices to the k power

  • Thread starter Dustinsfl
  • Start date
  • #1
699
5
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.
 

Answers and Replies

  • #2
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If [tex]B=P^{-1}AP[/tex], then [tex]B^k=[/tex]... ?
 
  • #3
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I forgot to mention I am proving this for each positive integer or the natural numbers.
 
  • #4
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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
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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
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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
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
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What should I do then with (S-1*A*S)k then?
 
  • #8
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Write it out.
 
  • #9
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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
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Now try it for (P-1AP)2=.....
I not sure about this part.
 
  • #11
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Just multiply the term by itself.
(XYZ)2=(XYZ)(XYZ)=XYZXYZ
 
  • #12
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Never mind my eyes were playing tricks on me I thought that was p- not to the power.
 
  • #13
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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
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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
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You will use induction after all. But now that you see the pattern, you should be able to make it work.
 
  • #16
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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
614
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(P-1AP)k+1 = (P-1AP)k(P-1AP). Then apply the induction hypothesis (assuming you proved a base case).
 
  • #18
699
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This is the step I am stuck on.
 
  • #19
614
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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.
 
  • #20
33,162
4,847
By assumption, Bk = P-1AkP.

So Bk+1 = BkB = ?
 
  • #21
699
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I was working the wrong side of the equation was the issue thanks.
 

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