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Math Help - Rank A and Basis of Rn

  1. #1
    Kyo
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    Rank A and Basis of Rn

    Hi guys, I would really apprieciate it if I could get some tips to proving this:

    If A is an m x n matrix with columns c1, c2, ... cn, and rank A = n, show that (I'll let B = A-transpose here)), {Bc1, Bc2, ... Bcn} is a basis of Rn.

    So far, I know that in order to show that it's a basis of Rn, it must span and be linearly independent.

    I know rank A = rank B = dim(col A) = dim(row A) = n

    I know that the columns of A are independent, therefore the rows of B are independent.
    The product of BA is then independent because it is invertible, and is the matrix [Bc1 Bc2 ... Bcn].
    I know the rows of A span Rn and therefore the columns of B span Rn.

    Am I wrong with assuming BA is independent due to invertibility? Since I think I will have to use this in the proof. And can I get a nudge towards the next step?

    Any help would be lovely, thank you!
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  2. #2
    MHF Contributor
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    Re: Rank A and Basis of Rn

    Hey Kyo.

    You can use the fact directly that det(A) <> 0 (and A is square) shows that all vectors are linearly independent and since det(A) = det(A^t) then A^t also has the same property.

    I'm not not sure how they expect you to assert det(A) = 0, but once you do that the rest follows.
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  3. #3
    Kyo
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    Re: Rank A and Basis of Rn

    Thanks chiro, but are you referring to AB being square? My A matrix is actually an m x n matrix.
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  4. #4
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    Re: Rank A and Basis of Rn

    Whatever matrix you are using as your basis (or trying to prove that a basis exists).
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