鹏程 张
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Facing a unbalanced matrix, such as 3*5000, why someone reduce the computational cost by svd(A*A') .
Computing singular value decomposition is the main computational cost in many algorithms . For a matrix A(m*n) ,if m is much la...
presque 3 ans il y a | 1 réponse | 0
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How does `svd(A*A')` reduce the computational cost?
Computing singular value decomposition is the main computational cost in many algorithms . For a matrixA(m*n) ,if m is much...
presque 3 ans il y a | 1 réponse | 0