What is the Geometric Interpretation of the Final Step in SVD?

[daviddoria]

<br /> Ax = U \Sigma V^T x<br />

(A is an m by n matrix)

I understand the first two steps,

1) V^T 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) \Sigma 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

 

[fresh_42]

The Wikipedia entry for it is quite long. It might contain already what you were looking for:
https://en.wikipedia.org/wiki/Singular_value_decomposition