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Math Help - crisis of identity (matrix transformations)

  1. #1
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    Smile crisis of identity (matrix transformations)

    Hi folks,

    I read the following example in my maths book and feel the authors have not explained what they are doing very clearly. We have to find the equations of any straight lines that are invariant under the following transformation:

     \left(  \begin{array}{c} x' \\ y' \end{array} \right) = <br />
\left( \begin{array}{cc} -6 & 3 \\ 4 & -2 \end{array} \right)<br />
\left( \begin{array}{c} x \\ y \end{array} \right) + <br />
\left( \begin{array}{c} 4 \\ -1 \end{array} \right)<br />

    Where x' and y' are the image points of x and y. Any straight line will be of the form y = mx + c and any point on the line will have coordinates of the form (k, mk + c).

    Thus

     \left(  \begin{array}{c} x' \\ y' \end{array} \right) = <br />
\left( \begin{array}{cc} -6 & 3 \\ 4 & -2 \end{array} \right) <br />
\left( \begin{array}{c} k \\ mk + c \end{array} \right) + <br />
\left( \begin{array}{c} 4 \\ -1 \end{array} \right)

    =

     \left( \begin{array}{c} -6k + 3mk + 3c + 4 \\ <br />
4k - 2mk - 2c - 1 \end{array} \right) <br />

    Now if y = mx + c is invariant, then this point (x', y') must also lie on y = mx + c

    i.e. 4k - 2mk - 2c - 1 = m(-6k + 3mk + 3c + 4) + c

    giving: k(4 + 4m - 3m^2) = 3mc + 4m + 3c + 1 (equation 1).

    This equation must be true for all values of k (i.e. equation 1 is an identity in k). Thus:

    4 + 4m - 3m^2 = 0 and  3mc + 3c + 1 = 0

    So: for those who have followed so far, I have to ask, how is equation 1 split into two parts? k is a variable and I know the difference between an equation and an identity (one is for all k, the other only particular values of k) but I don't understand the theory behind the way equation 1 suddenly becomes two equations both equal to zero. Can anyone help?

    best regards
    Last edited by s_ingram; February 23rd 2010 at 11:36 AM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by s_ingram View Post
    Hi folks,

    I read the following example in my maths book and feel the authors have not explained what they are doing very clearly. We have to find the equations of any straight lines that are invariant under the following transformation:

     \left(  \begin{array}{c} x' \\ y' \end{array} \right) = <br />
\left( \begin{array}{cc} -6 & 3 \\ 4 & -2 \end{array} \right)<br />
\left( \begin{array}{c} x \\ y \end{array} \right) + <br />
\left( \begin{array}{c} 4 \\ -1 \end{array} \right)<br />

    Where x' and y' are the image points of x and y. Any straight line will be of the form y = mx + c and any point on the line will have coordinates of the form (k, mk + c).

    Thus

     \left(  \begin{array}{c} x' \\ y' \end{array} \right) = <br />
\left( \begin{array}{cc} -6 & 3 \\ 4 & -2 \end{array} \right) <br />
\left( \begin{array}{c} k \\ mk + c \end{array} \right) + <br />
\left( \begin{array}{c} 4 \\ -1 \end{array} \right)

    =

     \left( \begin{array}{c} -6k + 3mk + 3c + 4 \\ <br />
4k - 2mk - 2c - 1 \end{array} \right) <br />

    Now if y = mx + c is invariant, then this point (x', y') must also lie on y = mx + c

    i.e. 4k - 2mk - 2c - 1 = m(-6k + 3mk + 3c + 4) + c

    giving: k(4 + 4m - 3m^2) = 3mc + 4m + 3c + 1 (equation 1).

    This equation must be true for all values of k (i.e. equation 1 is an identity in k). Thus:

    4 + 4m - 3m^2 = 0 and  3mc + 3c + 1 = 0

    So: for those who have followed so far, I have to ask, how is equation 1 split into two parts? k is a variable and I know the difference between an equation and an identity (one is for all k, the other only particular values of k) but I don't understand the theory behind the way equation 1 suddenly becomes two equations both equal to zero. Can anyone help?

    best regards
    If Ax+ B= Cx+ D for all x then, in particular, it is true for x= 0. Setting x= 0, the equation becomes B= D. But then our equation becomes Ax+ B= Cx+ B and, subtracting B from both sides, Ax= Cx. Again, since this is true for all x, it is true for x= 1. Setting x= 1, we have A= C.

    In this case, 4k - 2mk - 2c - 1 = m(-6k + 3mk + 3c + 4) + c or (4- 2m)k- (2c+1)= (-6m+ 3m^2)k+ (3mc+ 4m), A= 4-2m, B= -(2c+1), C= -6m+ 3m^2, and D= 3mc+4m. A= C gives 4- 2m,= -6m+3m^2 and B= D gives -(2c+1)= 3mc+4m.

    (Strictly speaking, here they are using a variation of this: if Ax+ B= 0 for all x, then A= 0 and B= 0. Here, C= D= 0.)

    More generally, if two polynomials are equal for all x then "corresponding coefficients", that is, coefficients of the same power of x, must be equal.
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  3. #3
    Member
    Joined
    May 2009
    Posts
    91

    Thumbs up

    Dear HallsofIvy,

    Thank you once again for a very clear and patient explanation! You must be a mathmatician!
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