Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric interpretation of SVD

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

    (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?

    Thanks!

    Dave
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted