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Change of Basis

  1. Apr 12, 2015 #1
    Suppose a change of basis from basis ##B## to basis ##C## is represented by the matrix ##S##.
    That is, ##S## is the transformation matrix from ##B## to ##C##.

    Now if ##t## is a given linear transformation, ##t:~V\rightarrow V##, with eigenvectors ##\epsilon_i##, say, and ##T## is the representation of ##t## in ##B##, then, the representation of t in ##C## is ##STS^{-1}##.

    Now, if the representation of ##\epsilon_i## in basis ##B## be ##v^B_i##, then the representation of ##\epsilon_i## in basis ##C##, ##v^C_i=Sv^B_i##.

    But shouldn't the vectors themselves transform in an opposite sense to the transformation of the basis, that is, shouldn't it be that ##v^C_i=S^{-1}v^B_i## ? I'm getting really confused. Please, can someone clarify?
     
  2. jcsd
  3. Apr 13, 2015 #2

    RUber

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    I agree with what you have here.
    Remember that S:B-->C. So what would it mean to take the inverse on something in B? As in:
    ##v^C_i=S^{-1}v^B_i##.
     
  4. Apr 14, 2015 #3

    Fredrik

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    Let's write ##B=(e_1,\dots,e_n)##, ##C=(f_1,\dots,f_n)##. Let ##T## be the linear operator such that ##Te_i=f_i## for all ##i##. I won't be writing any summation sigmas. There's always a sum over the index or indices that appear exactly twice. We have
    \begin{align}
    ([x]_B)_j e_j =x=([x]_C)_i f_i =([x]_C)_i Te_i = ([x]_C)_i (Te_i)_j e_j =([x]_C)_i ([T]_B)_{ji} e_j=([T]_B[x]_C)_j e_j.
    \end{align} This implies that ##[x]_B=[T]_B[x]_C##. If we define ##S=([T]_B)^{-1}##, we can write this as ##[x]_C=S[x]_B##, or equivalently as
    $$([x]_C)_i =S_{ij} ([x]_B)_j.$$ We also have
    $$f_i=Te_i =(Te_i)_j e_j =([T]_B)_{ji} e_j =(S^{-1})_{ji} e_j.$$
     
  5. Apr 14, 2015 #4

    WWGD

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    You know it's all about the base, about the base.
     
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