CODEWITHSUNDEEP
numpylinear-algebra
rando
rando

Reputation: 377

svd doesn't return correct dimension

I have a matrix with dimension (22,2) and I want to decompose it using SVD. SVD in numpy doesn't return the correct dimensions though.I'd expect dimensions like (22,22), (22),(22,2)?

enter image description here

Upvotes: 1

Views: 775

Answers (1)

darcamo
darcamo

Reputation: 3493

The returned dimensions are correct. The uu and vvh matrices are always square matrices, while depending on the software s can be an array with just the singular values (as in numpy) or a diagonal matrix with the dimension of the original matrix (as in MATLAB, for instance).

The dimensions of the uu matrix is the number of rows of the original matrix, while the dimension of the vvh matrix is the number of columns of the original matrix. This can never change or you would be computing something else instead of the SVD.

To reconstruct the original matrix from the decomposition in numpy we need to make s into a matrix with the proper dimension. For square matrices it's easy, just np.diag(s) is enough. Since your original matrix is not square and it has more rows than columns, then we can use something like

S = np.vstack([np.diag(s), np.zeros((20, 2))])

Then we get a S matrix which is a diagonal matrix with the singular values concatenated with a zero matrix. In the end, uu is 22x22, S is 22x2 and vvh is 2x2. Multiplying uu @ S @ vvh will give the original matrix back.

Upvotes: 2

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