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Rank question

  1. Mar 20, 2005 #1
    This is a T/F - prove type of question:

    A is m x n, M is matrix of TA with respect to bases B of R^m and B' of R^n. Then rank of A = rank of M.

    My reasoning is that it is true, since the lin. transf. is R^n->R^m, which means that in this formula:

    M = CB' A PB (CB' (coord matrix) is inverse of PB', and PB or PB' stands for point matrix with respect to the B or B' basis, respectively)

    PB has dimension n and PB' has dimension m, and CB' has the same dimension as PB' (inverting a matrix should not change dimension, right?). And it also should be true that both PB and CB' are square matrices (?) because they are invertible and we can invert only square matrices (?), so what I have is following composition of dimensions of the matrices in the formula for M:

    (m x m) (m x n) (n x n) = (m x n)

    Is this correct?
    Thanks in advance.
    Last edited: Mar 20, 2005
  2. jcsd
  3. Mar 21, 2005 #2

    matt grime

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    What is T? you've just introduced TA without explaining what it T is.
  4. Mar 21, 2005 #3
    Oh, sorry, T stands for linear transformation.
    Here is the full definition of the formula:
    M = CB' A PB,
    A is (m x n) matrix, B basis for R^n , B' basis for R^m, M matrix of TA: R^n->R^m with respec to basis B and B'.
  5. Mar 21, 2005 #4

    matt grime

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    Your question still confuses me a little.

    Can I suggest: let A be linear map from V to W, and T a linear marp from W to itself, Is the rank of A the same as the rank of TA?

    The answer is trivially no: let A be non-zero, and T be zero.

    Unless, by linear transformation you mean something other than what I take it to mean.
  6. Mar 21, 2005 #5


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    the rank of a matrix is the dimension of the image of the corresponding linear map. hence it makes no difference what basis you use to express it.
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