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Math Help - The rank-nullity theorem

  1. #1
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    The rank-nullity theorem

    Let T: R^n \rightarrow R^m be a linear transformation. Then there is a theorem that:

    RankT + nullity T = n

    This theorem is said to be equivalent to:

    dim(Ran(T))+ dim(Kernel(T))= n

    [n is the dimension of domain]

    So here's my question: how can we use the first one to derive/prove the second?

    This is my unsuccessful attempt:

    And for nullity(T):

    nullity(T) = dim(null(T))

    Since null(T) = ker(T)

    So, nullity(T) = dim(ker(T))

    For rank(T):

    rank(T) = dim(row(A))

    Since row(A) = null(A) and null(A) = ker(T)

    dim(row(A)) = dim(ker(T))

    ???



    I will be great if anyone could show me how to prove this. Thanks.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by Roam View Post
    Let T: R^n \rightarrow R^m be a linear transformation. Then there is a theorem that:

    RankT + nullity T = n

    This theorem is said to be equivalent to:

    dim(Ran(T))+ dim(Kernel(T))= n

    [n is the dimension of domain]

    So here's my question: how can we use the first one to derive/prove the second?

    For rank(T):

    ----------
    I will be great if anyone could show me how to prove this. Thanks.
    By definition, for a linear transformation T, dim(Range(T)) = Rank(T). Whats your definition?

    You seem to be confusing a linear transformation with its matrix(after choosing a basis).

    P.S: What is A?
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