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

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SUMMARY

The discussion centers on the geometric interpretation of the final step in Singular Value Decomposition (SVD), specifically how the matrix U transforms the output from the stretching operation performed by Σ back into R^m. The initial steps involve V^T rotating the input vector x in R^n, followed by Σ stretching the result. The confusion arises in understanding that while U does not serve as the inverse of V^T, it effectively maps the stretched vector into the original space of R^m, completing the transformation process.

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daviddoria
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<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
 
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