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anonymous

  • 5 years ago

Hey everyone..P, A, and B are nXn matrices and have the equation B=P^-1AP (so they are similar) Show that B^2=P^-1A^2P and find B^k and A^k

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  1. anonymous
    • 5 years ago
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    im new plz tell me what does ^ means

  2. anonymous
    • 5 years ago
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    It means "to the exponent of" so whatever comes before that symbol is raised to the exponent of whatever comes after that symbol

  3. anonymous
    • 5 years ago
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    Couldn't you just write, \[ B^{2} = (P^{-1} A P)(P^{-1} A P) \] and then use the fact that P and \(P^{-1}\) are inverses to write, \[ B^{2} = P^{-1} A (PP^{-1}) A P = P^{-1} A ( I ) A P = P^{-1} A^{2} P \] A similar arguement should get you a formula for \( B^{k} \). Once you have the formula for \( B^{k} \) you should be able to multiply that by P on the left and \( P^{-1} \) on the right you should get a formula for \( A^{k} \). Or at least you will if my quick thoughts on \( B^{k} \) are correct.

  4. anonymous
    • 5 years ago
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    Thanks a lot! Yes whatever B is raised to, A is raised to as well and the P's stay the same.

  5. anonymous
    • 5 years ago
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    Now I'm just working on finding B^k and A^k

  6. anonymous
    • 5 years ago
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    I haven't thought much more about it, but the formula for \( B^{k} \) should be just an extention of the \( B^{2} \).

  7. anonymous
    • 5 years ago
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    I mean the "proof" of the formula.....

  8. anonymous
    • 5 years ago
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    sorry just I haven't been great with proofs, I've got \[B ^{k}= B B ^{K-1} which -> =(P ^{-1}AP)(P ^{-1}AP)^{K-1}\] Then \[-> =(P ^{-1}A ^{2}P)^{K-1}\]...Is that sufficient proof? I don't know where to end it

  9. anonymous
    • 5 years ago
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    .....Sorry I think the answer is just \[B ^{k}=P^{-1}A ^{K}P\]...and that is sufficient

  10. anonymous
    • 5 years ago
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    And \[A ^{k}=P ^{-1}B ^{K}P\] correct? I'm not sure if we need more proof than that, it just says "find an analogous relationship involving B^k and A^k

  11. anonymous
    • 5 years ago
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    Sorry, I've been having internet issues....

  12. anonymous
    • 5 years ago
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    No problem, I am too

  13. anonymous
    • 5 years ago
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    You've got the correct formula for \(B^{k}\). I think the "best" proof (although that is relative) is just to do something like, \[ B^{k} = B B .... B B \] where you multiply the B k times then just replace each B with \( P^{-1} A P \) and cancel each \( P P^{-1} \) as above. The problem is more one of writing this out. It's more of a "thought" proof I suppose. It's probably going to depend on what you actually need to do. If It's just find a formula then you probably don't need a lot of proofs...

  14. anonymous
    • 5 years ago
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    okay Ill just do \[B ^{k}=BB ^{K-1}B ^{K-2}...B ^{K-K}\]

  15. anonymous
    • 5 years ago
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    For the \(A^{k}\) I think you've got the P's backwards. Remember that you've got to be careful with "order" of multiplication. I.e. you need to multiply on the same side of the equation. So, starting with the formula for \(B^{k}\) we can do the following \[ B^{k} = P^{-1} A^{k} P \] The multiply the left side by P and the right side by \(P^{-1}\). \[ P B^{k} P^{-1} = P P^{-1} A^{k} P P^{-1} \] which gives, \[ P B^{k} P^{-1} = A^{k} \]

  16. anonymous
    • 5 years ago
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    I'm going to be away from my computer for a while now unfortunately. I'll try to check back later, but I don't knwo when I'll get the chance....

  17. anonymous
    • 5 years ago
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    Thanks for everything i'm good now

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