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Linear Algebra - Change of Bases

  1. Mar 31, 2010 #1
    Hi, again another problem:

    Let B1 = {( [tex] \stackrel{1}{3}[/tex]),([tex] \stackrel{1}{2}[/tex])} and


    [tex]B_{2} = [ \frac{1}{\sqrt{2}}( \stackrel{1}{1}), \frac{1}{\sqrt{2}} (\stackrel{-1}{1}) ] [/tex]


    Determine the representing matrix [tex]T = K_{B_{2},B_{1}} \in \Re^{2\times2}[/tex] for the change from B1 coordinates to B2 coordinates.

    I have no idea what i should do here. I ve found how to calculate the representing matrix from a domain to a codomain.
    Is this the same way? Can you give me atleast a hint, please ^^.

    Thx Mumba

    PS. Sorry, it looks really strange. I dont know how to formate this better.
     
  2. jcsd
  3. Mar 31, 2010 #2
    This reminds me of quantum mechanics 1, a course that crushed m average and of which I remember very little.

    I tried a couple things and the answer i got was 1/sqrt(2) ( 2 5 )
    ( 0 1 )
    for the matrix T. But as i said again, thats just ag uess
     
  4. Mar 31, 2010 #3

    jbunniii

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    If S denotes the standard basis for [itex]\mathbb{R}^2[/itex], do you know how to find

    [tex]K_{S,B_1}[/tex] and [tex]K_{S,B_2}[/tex]?

    If so, then observe that

    [tex]K_{B_2,B_1} = K_{B_2,S} K_{S,B_1} = K_{S,B_2}^{-1}K_{S,B_1}[/tex]
     
  5. Mar 31, 2010 #4
    No, i dont even know what K is supposed to be...
     
  6. Mar 31, 2010 #5

    jbunniii

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    I am using your notation.

    [tex]K_{S,B_1}[/tex]

    is the change-of-basis matrix that transforms [itex]B_1[/itex] coordinates to [itex]S[/itex] coordinates. I suggested doing it this way because you already know how to express [itex]B_1[/itex] and [itex]B_2[/itex] in [itex]S[/itex] coordinates: that is what you are given.
     
  7. Mar 31, 2010 #6
    So i should calculate the change of base matrix for B1 to S and the inverse of the change of base matrix for B2 to S coordinates?
    And multiply this to get my result?

    Ok i ll try to find out how to calculate the change of base matrix ^^
    Thx
     
  8. Mar 31, 2010 #7
    Is [text]K_{S,B1}[/tex] not the same as B1?
    And the same for B2? So i calculate the inverste and multply them and then im finished?
     
  9. Mar 31, 2010 #8
    the i get as matrix:

    4/sqrt(2) 3/sqrt(2)
    2/sqrt(2) 1/sqrt(2)

    correct?
     
  10. Mar 31, 2010 #9

    jbunniii

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    Well, [itex]B_1[/itex] is a basis (set of vectors), not a matrix, so what you said is not exactly correct. However, I think what you are trying to say is this:

    [tex]K_{S,B_1}[/tex] is the matrix whose columns are the basis vectors from [itex]B_1[/itex] expressed in [itex]S[/itex] coordinates, namely

    [tex]K_{S,B_1} = \left[\begin{array}{cc}1 & 1 \\ 3 & 2\end{array}\right][/tex]

    and similarly for [tex]K_{S,B_2}[/tex].

    So yes, now you can find [tex]K_{B_2,B_1}[/tex] as I described earlier.
     
  11. Mar 31, 2010 #10
    Coool thx a lot, easy this way.
    :D
     
  12. Mar 31, 2010 #11

    jbunniii

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    I get the same answer.
     
  13. Apr 1, 2010 #12
    Cool :)
    Thx jbunniii
     
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