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Math Help - rank problem.....

  1. #1
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    rank problem.....

    Q:If A and B are 3 by 3 real matrices such that rank(AB)=1,then rank(BA) can't be
    a)0
    b)1
    c)2
    d)3
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  2. #2
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    Hint: Think about determinants (det(BA) = det(AB)).
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Hint: Think about determinants (det(BA) = det(AB)).
    plz clarify it more...because when i use ur hint i got two answers (3,2)
    but i have to select one.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Another big hint : if rank (AB)=1, \det (AB) cannot be ...?
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  5. #5
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    Quote Originally Posted by arbolis View Post
    Another big hint : if rank (AB)=1, \det (AB) cannot be ...?
    it can't be zero...is it right??
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Mathventure View Post
    it can't be zero...is it right??
    I thought so, but now I realize I'm confused and I think it's false. Disregard my post.
    EDIT: If rank (AB) is one, there are 2 null rows in the equivalent reduced matrix of AB, right? I have a conclusion but I don't know if it's true. I'll wait someone to help us.
    EDIT 2 : we were on the right track. det (AB)=0 since the its rank is 1. Hence det (BA) =0 and so it has at least a null row. What does it say about the rank?
    Last edited by arbolis; April 23rd 2009 at 06:50 PM.
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  7. #7
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    Quote Originally Posted by arbolis View Post
    I thought so, but now I realize I'm confused and I think it's false. Disregard my post.
    its okay...
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