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Math Help - Proving m=n in matrix problem

  1. #1
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    Proving m=n in matrix problem

    OK here is the problem
    Suppose \mathbf{A}\text{ is an }m\times n\text{ matrix and there exist matrices }\mathbf{C}\text{ and }\mathbf{D}\text{ such that }\mathbf{CA}=\mathbf{I}_n\text{ and }\\\mathbf{AD}=\mathbf{I}_m\text{. Prove that }m=n

    im soo clueless =( any help would be awesome
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  2. #2
    Forum Admin topsquark's Avatar
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    Here's a start. Let C be a p x m matrix. Then CA is a (p x m)(m x n) = p x n matrix. But CA = In. Thus C is an n x m matrix. Do this again for AD. What does all this mean for A?

    -Dan

    I fixed the problem. Unfortunately that makes my comment pointless.

    -Dan
    Last edited by topsquark; February 4th 2011 at 03:27 PM.
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  3. #3
    A Plied Mathematician
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    Reply to topsquark:

    Does that really work? I can see how you could get that C must be n x m, just to make the sizes work out. Why must it be n x n?

    I think you're going to have to use rank arguments here.
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  4. #4
    Forum Admin topsquark's Avatar
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    You are right. I had a small typo on the worksheet I scratched out the solution on. Apologies and thanks for the catch Ackbeet. 8]

    -Dan
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  5. #5
    A Plied Mathematician
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    The following facts may be useful:

    \text{rank}(I_{n})=n, and \text{rank}(I_{m})=m.

    Also, if A is an m\times n matrix, it is true that

    \text{rank}(A)\le\min(m,n).

    Furthermore, it is also true that

    \text{rank}(AB)\le\min(\text{rank}(A),\text{rank}(  B)).

    Using these facts, as well as facts about the minimum function, you should be able to prove both that m\le n, and that n\le m, thus proving your result.

    Can you see your way forward?
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