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A question in transformation

  1. Feb 7, 2008 #1
    i added a link with the question
    and a solution

    http://img518.imageshack.us/my.php?image=img8267rf4.jpg

    i want to transform a matrix from a standart basis to basis B
    A-out old matrix
    X- the old basis
    Y-the new basis
    S-the transformation matrix
    S^-1 -its inverse

    N-resolt
    i did
    Y|X>>>I|S

    S|I=I|S^-1

    N=S*A*S^-1

    in some metrices i got fractures so i multiplied by the dinominator the whole line

    that what i did
    i dont know where i did a mistake

    or wether this method is ok
     
  2. jcsd
  3. Feb 7, 2008 #2

    HallsofIvy

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    The method is ok and it looks like you did it correctly.
     
  4. Feb 7, 2008 #3
  5. Feb 10, 2008 #4
    i tried many times to solve it but i cant get nor to the formal answer nor
    to understand the formal solution

    can some solve it from the start to the end
     
  6. Feb 10, 2008 #5

    HallsofIvy

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    Sorry, I didn't look closely enough. You have, as the inverse to
    [tex] S^{-1}= \left(\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & 1 \end{array}\right)[/tex]
    [tex]S= \left(\begin{array}{ccc} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\ -\frac{1}{4} & -\frac{3}{4} & \frac{5}{4} \end{array}\right)[/tex]
    That is correct.

    However, in "S*A*S-1" you have removed the denominators! Why did you do that? I know you said "i got fractures so i multiplied by the dinominator the whole line" but that, of course, changes the matrix. You will need to go back and divide each line by that multiplier.
     
  7. Feb 10, 2008 #6
    i cant get to the answer of the solution that i was given

    i am not sure about how to build the S matrix

    (of transformation from one basis to the other)
     
  8. Feb 10, 2008 #7

    HallsofIvy

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    I told you that the S matrix is exactly what you said:
    [tex]S= \left(\begin{array}{ccc} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\ -\frac{1}{4} & -\frac{3}{4} & \frac{5}{4} \end{array}\right)[/tex]

    My question was "Why didn't you use that?"
     
  9. Feb 10, 2008 #8
    i did use that

    as you can see in my soltuion i multiplied the first one by 2

    the second one by 4
    and the third one by 4
    so as you you said i need to divide the resolt by these numbers

    i took the resolt
    and i tried to divide the rows by
    1/32
    but no way that whould give me the formal answer
     
    Last edited: Feb 10, 2008
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