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Math Help - proving matrix theorem

  1. #1
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    proving matrix theorem

    Hello,
    Can you please help solve the following problem:
    Given two matrices A,B and AB=BA, I need to prove that for every n:
    {(AB)}^{n}={A}^{n}{B}^{n}.

    Thanks in advance,
    Michael
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  2. #2
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    Quote Originally Posted by msokol89 View Post
    Hello,
    Can you please help solve the following problem:
    Given two matrices A,B and AB=BA, I need to prove that for every n:
    {(AB)}^{n}={A}^{n}{B}^{n}.
    Have you tried to prove this by induction?
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  3. #3
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    I suppose to technically prove this, you would use induction.

    But if you just write out the definition of exponentiation

    (AB)^n=ABAB\cdots AB ( n times)

    the result should be clear.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Have you tried to prove this by induction?
    I have tried to prove it by induction:
    1) I checked for n=1:
    {(AB)}^{1}=(A)}^{1}(b)}^{1}
    2)I assumed for n=k it is true:
     {(AB)}^{k}=(A)}^{k}(b)}^{k}
    3)I checked for n=k+1:
    {(AB)}^{k+1}=(A)}^{k+1}(b)}^{k+1}
    {(AB)}\cdot {(AB)}^{k}={A}\cdot {A}^{k}\cdot {B}\cdot {B}^{k}
    Then, after switching places and using the induction assumption , I reached:
    {(AB)}\cdot {(AB)}^{k}={(AB)}\cdot {(AB)}^{k}

    is this ok?
    Last edited by msokol89; March 9th 2011 at 09:15 PM.
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  5. #5
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    Quote Originally Posted by msokol89 View Post
    I have tried to prove it by induction:
    1) I checked for n=1:
    {(AB)}^{1}=(A)}^{1}(b)}^{1}
    2)I assumed for n=k it is true:
     {(AB)}^{k}=(A)}^{k}(b)}^{k}
    3)I checked for n=k+1:
    {(AB)}^{k+1}=(A)}^{k+1}(b)}^{k+1}
    {(AB)}\cdot {(AB)}^{k}={A}\cdot {A}^{k}\cdot {B}\cdot {B}^{k}
    Then, after switching places and using the induction assumption , I reached:
    {(AB)}\cdot {(AB)}^{k}={(AB)}\cdot {(AB)}^{k}

    is this ok?
    I think the last couple steps are... out of order or something.

    You have

    (AB)^{k+1}=(AB)(AB)^k=ABA^kB^k=AA^kBB^k=A^{k+1}B^{  k+1}.

    Your last line is {(AB)}\cdot {(AB)}^{k}={(AB)}\cdot {(AB)}^{k}. But... this is clearly true, it doesn't say anything about.. anything.
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  6. #6
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    it proves that this is true for n=k+1, therefore the entire statement is true.
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  7. #7
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    Quote Originally Posted by msokol89 View Post
    it proves that this is true for n=k+1, therefore the entire statement is true.
    No, it doesn't. You need to prove that (AB)^{k+1}= (AB)(AB)^k= A^{k+1}B^{k+1}. The trivial observation that (AB)(AB)^k= (AB)(AB)^k doesn't do t

    Having said that (AB)^k= A^kB^k, you can conclude that (AB)^{k+1}= (AB)(AB)^k= (AB)(A^kB^k). Now, what can you conclude from that?
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    No, it doesn't. You need to prove that (AB)^{k+1}= (AB)(AB)^k= A^{k+1}B^{k+1}. The trivial observation that (AB)(AB)^k= (AB)(AB)^k doesn't do t

    Having said that (AB)^k= A^kB^k, you can conclude that (AB)^{k+1}= (AB)(AB)^k= (AB)(A^kB^k). Now, what can you conclude from that?
    ok thanks,
    Do you have a way of solving the problem?
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  9. #9
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    Quote Originally Posted by msokol89 View Post
    Hello,
    Can you please help solve the following problem:
    Given two matrices A,B and AB=BA, I need to prove that for every n:
    {(AB)}^{n}={A}^{n}{B}^{n}.
    It looks as though you need to use induction twice if you really want to pin this result down rigorously.

    First, use induction to show that AB^n = B^nA for every n\geqslant1.

    Then use that result to prove by induction on m that A^mB^n = B^nA^m for every m\geqslant1 (and hence in particular when m=n).
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  10. #10
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    Quote Originally Posted by Opalg View Post
    It looks as though you need to use induction twice if you really want to pin this result down rigorously.

    First, use induction to show that AB^n = B^nA for every n\geqslant1.

    Then use that result to prove by induction on m that A^mB^n = B^nA^m for every m\geqslant1 (and hence in particular when m=n).
    I understand, thanks.
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