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Thread: Proving m=n in matrix problem

  1. #1
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    Proving m=n in matrix problem

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

    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; Feb 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:

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

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

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

    Furthermore, it is also true that

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

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

    Can you see your way forward?
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