Geometric interpretation of SVD

  • Thread starter daviddoria
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[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
 

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