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Homework Help: Matrix Transformation

  1. Apr 1, 2008 #1
    Hiya, just a quick question regarding matrices:

    The following 2 column vectors are a particular form of transformation when applied to 2x2 matrices:

    (1) (0)
    (0) (1)

    Am I right in saying the first, takes just the x component of the matrice, and the second the y component? Thing is, To me that sounds odd because the matrice's given are just matrices, with nothing to indicate they may be vectors or whatever, they seem to be just arrays of numbers.
  2. jcsd
  3. Apr 1, 2008 #2
    You're thinking too hard. [1 0; 0 1] x [a b; c d] = ? (I've delimited rows by ; )
  4. Apr 2, 2008 #3
    If I do what you wrote, you just get the same matrix again, but if you do it by my column vectors you get only half the matrix..
  5. Apr 3, 2008 #4
    The matrix M={{1,0},{0,1}} is an identity matrix. When it multiplies a matrix
    N={{a,b},{c,d}} the result is M N=N. In this case you can interpret the rows of the
    identity matrix as unit vectors i, j in the x and y directions respectively. Then M N takes the dot product of i and j with the columns of N, thus resulting in the original N.
  6. Apr 5, 2008 #5
    Ah i see, what I meant is, (1, 0 ) being one vector, which is then multiplied by M. ie. M.I, which gives just a rotation through i? I can see when my 2 column vectors are combined they equal an identity matrix, however I only want, one of the columns at any time, to see what they do to my matrix.
    M={1,0}, N={{a,b},{c,d}} mn= a,b ... what actually is this called? ie. its only part of the original matrix, I think the answer given is, its a rotation of some sort, along the x axis.
  7. Apr 5, 2008 #6
    If I understand the question as first stated, you are asking for an
    interpretation of the results of the matrix multiplication I N, where
    1 0
    0 1

    and N=

    a b

    The result of applying the rules of matrix multiplication is I N=N, that is why I is called
    an identity matrix.

    As for an interpretation of the multiplication, the dot product of the first row of I
    with the first column of N gives a, and so on.

    Incendentially, there is no rotation here.
  8. Apr 5, 2008 #7
    MN is not a valid matrix product, unless you meant MTN. NM is also a valid matrix product. The map f(N) = NM just selects the first column of the matrix N, the image of the first basis vector. The map f(N) = MTN maps N to the image of the first basis vector of the dual space in the dual of the image of N. If N is a rotation matrix, you will note that the result is a rotation of the basis vector.
    Last edited: Apr 5, 2008
  9. Apr 6, 2008 #8
    AH, nope sorry, I wasn't being clear, I mean, I have 2 different column vectors, (which happen to look just liike an identity matrix when put together)
    Slider142 's explanation is kinda of what i'm looking for, except I don't quite understand it.
    I did mean:(1 ). M. By mapping image, do you mean, the vector is placed along 1 ?
    (0 ) 0
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