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Math Help - Rank and Dim of the image - relation?

  1. #1
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    Rank and Dim of the image - relation?

    Hi - Need help in understanding the following

    Let
    X be a n-dim vector space
    Y be a m-dim vector space
    A be a mxn matrix

    Thus, AX defined a linear transformation,T from X into Y i.e. T: X -> Y
    Image of X under T, T(X) is a subspace of Y

    I need help in understanding and thus formally proving two concepts:

    1. Rank of A (which say is defined as dim of row space or dim of col space) = dim of T(X)

    2. Why is dim(row space of A) = dim(col space of A)

    Any pointers would be welcome. Thanks
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    Hi - Need help in understanding the following

    Let
    X be a n-dim vector space
    Y be a m-dim vector space
    A be a mxn matrix

    Thus, AX defined a linear transformation,T from X into Y i.e. T: X -> Y
    Image of X under T, T(X) is a subspace of Y

    I need help in understanding and thus formally proving two concepts:

    1. Rank of A (which say is defined as dim of row space or dim of col space) = dim of T(X)

    2. Why is dim(row space of A) = dim(col space of A)

    Any pointers would be welcome. Thanks
    For 1, Consider first a linear transformation T:\mathbb{Re}^4 \rightarrow \mathbb{Re}^4 associated to a 4 by 4 matrix A. Then, A has four columns. If three columns of them form a basis for a column space of A, the image of the corresponding matrix transformation spans 3-dimensional subspace of \mathbb{Re}^4.

    Now consider a linear transformation T:\mathbb{Re}^n \rightarrow \mathbb{Re}^m associated to a m by n matrix A. Then the column space of A consists of all vectors in \mathbb{Re}^m that are images of at least one vectors in \mathbb{Re}^n under matrix multiplication by A. If k columns of A form a basis for a column space of A, then the image of the corresponding matrix transformation spans k-dimensional subspace of \mathbb{Re}^m, where k is a positive integer less than or equal to m.

    For 2, you can use the fact that if A is any matrix, then rank (A) = rank (A^T).
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