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Change of coordinate matrix

  1. Jul 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove: Let [tex] A \in \mathrm{M}_{n \times n}(\mathbb{F}) [/tex] and let [tex] \gamma [/tex] be an ordered basis for [tex] \mathbb{F}^n [/tex]. Then [tex] [\boldmath{L}_A]_{\gamma} = Q^{-1}AQ [/tex], where Q is the nxn matrix whose jth column is the jth vector of [tex] \gamma [/tex].

    2. Relevant equations
    [tex] \boldmath{L}_A [/tex] denotes the left-multiplication transformation.

    3. The attempt at a solution
    Let [tex] \beta [/tex] be the standard ordered basis for [tex] \mathbb{F}^n [/tex] and C the change of coordinate matrix from [tex] \beta [/tex]-coordinates to [tex] \gamma [/tex]-coordinates. Then [tex] [\boldmath{L}_A]_{\beta} = A [/tex] and we have [tex] [\boldmath{L}_A]_{\gamma} = C^{-1}AC [/tex]. Where I'm stuck is showing that the jth column of C is the jth vector of [tex] \gamma [/tex]. Any hints would be appreciated.
  2. jcsd
  3. Jul 22, 2009 #2


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    The matrix C whose jth column is the jth vector of gamma maps the standard basis (1,0...0), (0,1,...0)...(0,0...1) into gamma, doesn't it?
  4. Jul 23, 2009 #3
    Dick, thanks for the response. I think I got it now. C should be the change of coordinate matrix from gamma-coordinates to beta-coordinates, not beta to gamma as I had stated above at first. In other words, for [tex] [\boldmath{L}_A]_{\gamma} = C^{-1}AC [/tex] to hold true, [tex] C = [\boldmath{I}]_{\gamma}^{\beta} [/tex] where I is the identity transformation. Then everything makes sense.
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