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Change of basis theorem?

  1. Dec 14, 2012 #1
    Let B={b1,b2} and C={c1,c2} be basis. Then the change of coordinate matrix P(C to B) involves the C-coordinate vectors of b1 and b2. Let
    [b1]c=[x1] and [b2]c=[y1]
    .........[x2]................[y2].

    Then by definition [c1 c2][x1]=b1 and [c1 c2][y1]=b2. I dont get how you can
    ................................. [x2].....................[y2]
    multiply the matrix with basis set C with the change of coordinate matrix P(C to B) to get back basis set B ?
    Can anyone help me understand how the derive the fact that you can take the set C basis and matrix P to get basis b1 and b2? My textbook just says very little about it.

    Here is an example of a problem relating to this idea.
    There was a problem that stated find a basis {u1,u2,u3} for R^3 such that P is the change of coordinates matrix from{u1,u2,u3} to the basis {v1,v2,v3}? P was given and v1,v2,v3were given as well. I know how to do it but don't get the how it works?
     
    Last edited: Dec 15, 2012
  2. jcsd
  3. Dec 19, 2012 #2
    Ok. Lets work in Two Dimensions because writing math text on this forum is slow for me.

    Suppose there is a vector v = <x1,x2>. We are used to the regular cartesian basis vectors
    [itex]{e}_1[/itex] = <1,0> and [itex]{e}_2[/itex] = <0,1> such that

    v = [itex]<x1,x2>^{T}[/itex] = x1[itex]{e}_1[/itex] + x2[itex]{e}_2[/itex]

    now lets say we want to express v in a new basis like [itex]{f}_1[/itex] = <1,1> and [itex]{f}_2[/itex] = <1,2> such that

    v = [itex]<y1,y2>^{T}[/itex] = y1[itex]{f}_1[/itex] + y2[itex]{f}_2[/itex]

    since v is the same vector regardless of how it is represented, we can equate the two basis expressions

    v = v

    y1[itex]{f}_1[/itex] + y2[itex]{f}_2[/itex] = x1[itex]{e}_1[/itex] + x2[itex]{e}_2[/itex]

    <[itex]{f}_1[/itex],[itex]{f}_2[/itex]>[itex]<y1,y2>^{T}[/itex] = <[itex]{e}_1[/itex],[itex]{e}_2[/itex]>[itex]<x1,x2>^{T}[/itex]

    Let <[itex]{f}_1[/itex],[itex]{f}_2[/itex]> = [itex]F[/itex] (2x2 Matrix with f basis as columns)
    Let <[itex]{e}_1[/itex],[itex]{e}_2[/itex]> = [itex]E[/itex] (2x2 Matrix with f basis as columns)

    Then

    [itex]F[/itex][itex]<y1,y2>^{T}[/itex] = [itex]E[/itex][itex]<x1,x2>^{T}[/itex]

    So

    [itex]<y1,y2>^{T}[/itex] = [itex]F^{T}[/itex][itex]E[/itex][itex]<x1,x2>^{T}[/itex]

    Let [itex]P[/itex] = [itex]F^{T}[/itex][itex]E[/itex]

    Then

    [itex]<y1,y2>^{T}[/itex] = [itex]P[/itex][itex]<x1,x2>^{T}[/itex]

    I hope this shows where the logic is coming from. If it is still not clear, let me know, and I will clarify. I will edit this better when I return home.
     
  4. Dec 19, 2012 #3
    thanks this is what I was looking for. I get it.
     
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