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A question in basis transformation

  1. Feb 28, 2008 #1
    Last edited: Feb 28, 2008
  2. jcsd
  3. Feb 28, 2008 #2

    HallsofIvy

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    You could check it your self, In terms of the standard basis, Ttr(1, 1, 1)= (3, 3, 7), Ttr(1, 0, 0)= (2, 1, 1), and Ttr(0, 0, 1)= (0, 0, 5). In terms of the B basis, those results would be (3, 3, 7)= a(1, 1, 1,)+ b(1, 0, 0)= c(0, 0, 1)= (a+ b, a, a+ c) so we have a+ b= 3, a= 3, a+ c= 7 which gives a= 3, b= 0, c= 4 or <3, 0, 4> (I am using "< >" for vectors written in the B basis). Similarly, (2, 1, 1) gives a= 1, b= 1, c= 0 or <1, 1, 0> and (0, 0, 5) gives < 0, 0, 5>. If you try to do everything in the "B" basis: multiply your "TB" matrix by <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> you get the first, second, and third columns, respectively. And they are NOT the same.

    You error was when you formed the "S-1" transformation matrix: you used the B basis vectors as rows and they should be columns. Use
    [tex]\left(\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 1\end{array}\right)[/tex]
    instead and you should be alright.
     
    Last edited: Feb 28, 2008
  4. Feb 28, 2008 #3
    i am confused about the vector apeareance
    i what form should i put it in the matrix
    as a row

    i what form should i put it in the matrix
    as a column


    ???

    also i was tald that in transformation we put the vectors as rows

    for example:
    in this sort of question i was to find the basis of V

    http://img301.imageshack.us/my.php?image=img83241re6.jpg

    first i thought that when a vector is signed as (x,y,z)
    we flip him verticaly
    and when its
    (x)
    (y)
    (z)
    then it should flip it horisontaly
    but apparently thats not how it works

    how it works??
    how do i write the given vectors in the metrix
    and in what form and in what cases??
     
    Last edited: Feb 28, 2008
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