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Geometric interpretation of SVD

  1. Sep 22, 2008 #1
    Ax = U \Sigma V^T x

    (A is an m by n matrix)

    I understand the first two steps,

    1) [tex]V^T[/tex] takes x and expresses it in a new basis in R^n (since x is already in R^n, this is simply a rotation)

    2) [tex]\Sigma[/tex] takes the result of (1) and stretches it

    The third step is where I'm a bit fuzzy...

    3) U takes the result of (2) and puts it into R^m. In eigenvalue decomposition, this is just the inverse transformation of V^T, but I always read "this is not the inverse transformation of step 1"

    Can someone clarify this last step a bit for me?


  2. jcsd
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