How does `svd(A*A')` reduce the computational cost?
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Computing singular value decomposition is the main computational cost in many algorithms .
For a matrixA(m*n) ,if m is much larger than n , one can compute the SVD of A*A',and then get an approximate SVD of by simple operations to reduce the computational cost.
How does it reduce the computational cost?
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Cutie
2021-6-21
SVD reduces computational costs because it provides a numerically stable matrix decomposition. You may refer to https://www.youtube.com/playlist?list=PLMrJAkhIeNNSVjnsviglFoY2nXildDCcv for detailed wokrings of SVD
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