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Math Help - Help! Proof of Rank A transpose = Nul A perp.

  1. #1
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    Help! Proof of Rank A transpose = Nul A perp.

    Prove.

    Rank(A^T ) = (Null(A))^\perp

    Where, if V is a subspace,

    V^\perp  = \{\vec{w} : \vec{w} is perpendicular to  \\\vec{v} \ \ \forall  \ \ \vec{v} \ \ \varepsilon \ \  V \}


    Now, I know the rank is number of linearly independent columns in a matrix. So the rank of A^T would be the number of linearly independent rows of A.

    I know that the nullspace is the solutions to A\vec{x}=\vec{0}
    However, the "perpendicular compliment" is a term I am not familiar with. And the simple definition does not really help me. Can someone help me with this proof?
    Last edited by Lovemachinzero; January 25th 2010 at 08:10 PM.
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  2. #2
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    Quote Originally Posted by Lovemachinzero View Post
    If A is a subspace of a vectorspace, prove that

    Rank(A^T ) = Null(A)^\perp

    Where, if V is a subspace,

    V^\perp  = \{\vec{w} : \vec{w} is perpendicular to  \\\vec{v} \ \ \forall  \ \ \vec{v} \ \ \varepsilon \ \  V \}


    Now, I know the rank is number of linearly independent columns in a matrix. So the rank of A^T would be the number of linearly independent rows of A.
    But you started by saying that A is a subspace of some vector space, not a matrix! And if A is a matrix, I don't know what you mean by " Null(A)^\perp".


    I know that the nullspace is the solutions to A\vec{x}=\vec{0}
    However, the "perpendicular compliment" is a term I am not familiar with. And the simple definition does not really help me. Can someone help me with this proof?
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  3. #3
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    I apologize. A is a matrix, and the perp should be inclusive of the subspace Null(A). I edited the first post!

    Now, I think that since Null(A) is the set of vectors \vec{x} such that A\vec{x}=\vec{0} The perpendicular compliment of Nul(A) would mean that there is a set of vectors \vec{v} such that <\vec{v},\vec{x}> = 0

    Is the Range(A^T) = Col(A^T) = Row(A)? Am I even on the right track?! =]
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