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Change of basis

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Recall that the matrix for [tex]T: R^{2} \rightarrow R^{2}[/tex] defined by rotation through an angle [tex]\theta[/tex] with respect to the standard basis for [tex]R^{2}[/tex] is

    [tex]\[A =\begin{array}{cc}cos \theta & -sin \theta \\sin \theta & cos\theta \\\end{array}\]\right][/tex]

    a) What is the matrix of T with respect to the basis (2,1),(1,-2)?

    2. Relevant equations



    3. The attempt at a solution
    So, here's the deal, I missed this particular lecture last week. I'm trying to teach myself this concept of changing basis. Here's one of my homework problems. Can you please check it?

    alpha = standard basis, beta = new basis

    I(1,0) = a(2,1) + b(1,-2) --> a = .4, b = .2
    I(0,1) = a(2,1) + b(1,-2) --> a = .2, b = -.4

    [tex]\[[I^{\beta}_{\alpha}] = \begin{array}{cc} .4 & .2 \\ .2 & -.4 \\ \end{array}[/tex]

    [tex]\[[I^{\beta}_{\alpha}]^{-1} = (\frac{1}{.4(-.4)-.2(.2)})\begin{array}{cc} -.4 & -.2 \\ -.2 & .4 \\ \end{array}
    = -5(\begin{array}{cc} -.4 & -.2 \\ -.2 & .4 \\ \end{array})
    = \begin{array}{cc} 2 & 1 \\ 1 & -2 \\ \end{array}[/tex]

    The new transformation matrix should be [tex][I^{\beta}_{\alpha}]^{-1}A[I^{\beta}_{\alpha}] = (\begin{array}{cc} 2 & 1 \\ 1 & -2 \\ \end{array})(\begin{array}{cc}cos \theta & -sin \theta \\sin \theta & cos\theta \\\end{array})(\begin{array}{cc} .4 & .2 \\ .2 & -.4 \\ \end{array}) = \begin{array}{cc}cos \theta & sin \theta \\ -sin \theta & cos \theta[/tex]
     
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  3. Mar 29, 2010 #2

    vela

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    Your solution is almost correct. You just got the two transformation matrices backwards.

    I just remember the order like this. Your I matrix (not the best letter to use) converts coordinates from the standard basis to the new basis, and its inverse goes from the new basis to the standard basis. The matrix A works on vectors in the standard basis. So suppose you have a vector x expressed in the new basis and you want to see what T does to it. First, you multiply I-1 to convert it to the standard basis, giving you I-1x. Once it's expressed in the standard basis, you can use A to calculate what T does to it, so now you have AI-1x. But A gives you a result that still in the standard basis, so you need to convert back to the new basis. The end result is T(x)=IAI-1x.
     
  4. Mar 29, 2010 #3
    You're right, I is not the best letter to use. I think my professor used it for "identity mapping".

    Anyways, I looked at a half-example from my teacher's notes. (She writes everything down on a smart board and saves it.) She put alpha = new basis and beta = standard basis and I got confused.
     
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