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Finding the range of a matrix

  1. Feb 27, 2014 #1
    1. The problem statement, all variables and given/known data
    ##\begin{bmatrix}1 & 1 & -1 & -1 \\1 & 2 & 0 & 1 \\-1 & 1 & 3 & 5 \\2 & 3 & -1 & 0\end{bmatrix}##
    a) Determine the range of L_A

    2. Relevant equations
    None

    3. The attempt at a solution
    The row-reduced matrix is as follows
    ##\begin{bmatrix}1 & 0 & -2 & -3 \\0 & 1 & 1 & 2 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{bmatrix}##

    Then
    ##2x_{3}+3x_{4}##

    ##-x_{3}-2x_{4}##

    ##x_{3}##

    ##x_{4}##

    Is this correct?



    1. The problem statement, all variables and given/known data
    ##L \big( \begin{bmatrix}a & b \\c & d \end{bmatrix} \big) = \begin{bmatrix}0 & a \\b & c \end{bmatrix}##
    a) Show that L is a linear transformation.
    b) Define L^{k} = L \circ L^{k-1} for every integer k >= 2

    2. Relevant equations
    To be a linear transformation, these must be true
    i) ##f(x_{1})+f(x_{2})=f(x_{1} + x_{2})##
    ii) ##cf(x_{1})=f(cx_{1})##

    3. The attempt at a solution
    a) I'm not sure how to start showing this. For i) do I add the two matrices? For ii) do I just multiply each entry of the first matrix by c?
    b) I don't know where to start at all. I'm not even sure what the question is asking.
     
    Last edited: Feb 27, 2014
  2. jcsd
  3. Feb 27, 2014 #2

    BvU

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    Is this two separate exercises ? Then why not start two separate threads?

    As for the second exercise:
    i) yes, but I would not call them "the" two matrices, but "any" two matrices
    ii) yes, but you confuse everyone if you use c twice.

    For the first exercise I think you want to check what values linear combinations of the columns can assume. So why row-reduce the thing ?
     
  4. Feb 27, 2014 #3
    For the first exercise, I though to row reduce it then find that x3 and x4 can be any real number by x1 and x2 depend on x3 and x4.

    For the second exercise, my textbook uses that same notation and uses c twice.
     
  5. Feb 27, 2014 #4

    BvU

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    Row reducing is fine (I did some reading up...) because it doesn't affect the dependence relations between the column vectors.
    This way you find that first two columns are linearly independent. 3 and 4 are linear combinations of these two.

    Don't use the same c. Prove that ##L \big( x \begin{bmatrix}a & b \\c & d \end{bmatrix} \big) = x\ L \big( \begin{bmatrix}a & b \\c & d \end{bmatrix} \big) ##
     
  6. Mar 1, 2014 #5
    So how does that lead me to finding the range?
     
  7. Mar 2, 2014 #6
    Bump.
     
  8. Mar 2, 2014 #7

    micromass

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    Take the standard basis ##\{e_1,e_2,e_3,e_4\}## of ##\mathbb{R}^4##. This is a basis so it spans the entire space.
    Now, we know that the image ##\{L_A(e_1), L_A(e_2), L_A(e_3), L_A(e_4)\}## spans the range of ##L_A##. So find the images of the basis vectors to obtain a set that spans the space. Then see if you can extract a basis from the set.
     
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