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Matrix to the power t

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data
    If A= [2 4 6 8] find AT


    2. Relevant equations



    3. The attempt at a solution
    Would I simply distribute the T across the matrix like this: [2T 4T 6T 7T 8T]
     
  2. jcsd
  3. Dec 9, 2012 #2

    Mute

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    The superscript 'T' here indicates the transpose of the matrix A. For a matrix A, if the element in the ith row and jth column is ##A_{ij}##, then for the transpose matrix ##A^T##, the element in the ith row and jth column is ##A_{ji}##.

    Note that if A is an mxn matrix, ##A^T## is an nxm matrix.
     
  4. Dec 9, 2012 #3
    Okay, thank you so much!
     
  5. Dec 9, 2012 #4

    Ray Vickson

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    Ia A a true matrix? What you have written is a 4-dimensional row vector. Its transpose would be the 4-dimensional column vector with elements 2,4,5,8 in that order; that is:
    [tex] [2\;4\;6\;8]^T = \left[ \begin{array}{c}{2\\4\\6\\8}\end{array}\right].[/tex]
     
  6. Dec 9, 2012 #5
    It is given to me by my teacher as a matrix, but I guess if it's a vector than it is a vector.

    I did transpose it that way tho :D

    Thanks!
     
  7. Dec 10, 2012 #6

    Ray Vickson

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    A vector is a special case of a matrix. However, perhaps you were supposed to interpret A as a 2x2 matrix, so A = [[2,4],[6,8]], where these are the two rows; that is:
    [tex] A = \pmatrix{2&4\\6&8}.[/tex]
    The transpose of A is
    [tex] A^T = \pmatrix{2 & 6 \\4 & 8}.[/tex]

    On the other hand, A to the power t is considerably more complicated. It would take several pages to write out the answer.
     
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