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I Change of basis question

  1. Sep 6, 2016 #1
    I have two n-vectors [itex]e_1, e_2[/itex] which span a 2D subspace of [itex]R^n[/itex]:
    [tex]
    V = span\{e_1,e_2\}
    [/tex]
    The vectors [itex]e_1,e_2[/itex] are not necessarily orthogonal (but they are not parallel so we know its a 2D and not a 1D subspace). Now I also have a linear map:
    [tex]
    f: V \rightarrow W \\
    f(v) = A v
    [/tex]
    where [itex]A[/itex] is a given [itex]n \times n[/itex] invertible matrix.

    My question is: how would I construct an orthonormal basis for the space [itex]W[/itex]?

    My thinking is to perform a QR decomposition on the [itex]n \times 2[/itex] matrix
    [tex]
    \left(
    \begin{array}{cc}
    A e_1 & A e_2
    \end{array}
    \right)
    [/tex]
    and then the columns of [itex]Q[/itex] will be an orthonormal basis for [itex]W[/itex]. Is this a correct solution? I'm not entirely sure since [itex]e_1,e_2[/itex] are not orthonormal.
     
  2. jcsd
  3. Sep 6, 2016 #2
    How do you define ##W##? Is ##W## the image of ##V## under the map ##A## (as a linear operator in ##R^n##)? In this case ##W## is two dimensional if ##A e_1## and ##A e_2## are linearly independent, and is one dimensional otherwise.

    In the case where ##W## is two dimensional, you know ##A e_1## and ##A e_2## form a basis. Use the Gram-Schmidt process to find orthonormal basis.
     
    Last edited: Sep 6, 2016
  4. Sep 6, 2016 #3
    Yes, we can think of [itex]W[/itex] as the image of [itex]V[/itex] under the map [itex]A[/itex]
     
  5. Sep 8, 2016 #4

    chiro

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