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Linearly Independent R^3 -> Linearly Independent R^4

  1. Feb 2, 2010 #1
    Standard Matrix for R^3 -> R^4 where if domain is Linearly Independent so is codomain

    1. The problem statement, all variables and given/known data

    5. Suppose T : R^3 -> R^4 is a linear transformation with the following property:
    For any linearly independent vectors v_1, v_2 and v_3 in R^3, the images T(v_1),
    T(v_2) and T(v_3) are linearly independent in R4.
    (a) Give an example of such a linear transformation. Give its standard matrix and the
    reduced row echelon form of this matrix.
    (b) Work out all possible shapes of the reduced row echelon form for such a matrix.
    Use the symbols 0, 1, and * where * indicates an entry which may take on the value
    of any real number.

    2. Relevant equations

    No relevant equations. The linear transformation must be given as a standard matrix with dimensions 3 * 4 since that would lead to a vector in R^4.

    3. The attempt at a solution

    So far I know that since the vectors are linearly independent, the reduced form of both vectors must have pivots in each column. I reasoned that the function must be injective because each vector in R^3 can only correspond to 1 vector in R^4 but there are more vectors in R^4 than R^3.

    This question seems confusing and I can't put my thoughts together. This is not a very advanced class (it is just a standard freshman linear algebra course for scientists and scientists).
    Last edited: Feb 2, 2010
  2. jcsd
  3. Feb 3, 2010 #2


    User Avatar
    Homework Helper

    If v = (a, b, c), try T(v) = (a, b, c, 0).
  4. Feb 3, 2010 #3
    Thank you for your response. After more thinking I was able to find that the matrix must be linearly independent (pivot in every column) and that when it is row reduced, the matrix would be:

    [ 1 0 0 ]
    [ 0 1 0 ]
    [ 0 0 1 ]
    [ 0 0 0 ]
  5. Feb 3, 2010 #4


    Staff: Mentor

    Don't you means that the columns in the matrix must be linearly independent?
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