Why singular value decomposition and non-negative matrix factorization gives unphysical results?
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For a physical process involving the change in populations of multiple spectral species, one can obtained a bi-linearly degenerate matrix.
It can be decomposed as the outer product of spectral species (column vectors) and kinetics (row vectors), where the number of vectors is the rank of the matrix.
It is known that the column vectors are positive, and it is physically required that the row vectors to be positive as well (negative population is not possible).
However, when performing SVD, it gives sign-changing vectors for both left and right singular vectors.
When performing NNMF, where the vectors are constrained to be positive and the orthogonality loosely holds. The 'left vectors' result of NNMF looks like the left singular vectors in SVD, but either 'shifted up' or have their negative elements replaced by 0.
Assuming the rank of 3, the column vectors it gives are [A B C],which are all positive but still not physical. The physical solution takes the form of [A B iA+jC]. So C is actually somewhat the difference between A and the 'actual' C.
Is it possible to obtain the physical solution by mixing the left vectors ?
linear-algebra matrices matrix-decomposition svd
asked Aug 14 at 9:49
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For a physical process involving the change in populations of multiple spectral species, one can obtained a bi-linearly degenerate matrix.
It can be decomposed as the outer product of spectral species (column vectors) and kinetics (row vectors), where the number of vectors is the rank of the matrix.
It is known that the column vectors are positive, and it is physically required that the row vectors to be positive as well (negative population is not possible).
However, when performing SVD, it gives sign-changing vectors for both left and right singular vectors.
When performing NNMF, where the vectors are constrained to be positive and the orthogonality loosely holds. The 'left vectors' result of NNMF looks like the left singular vectors in SVD, but either 'shifted up' or have their negative elements replaced by 0.
Assuming the rank of 3, the column vectors it gives are [A B C],which are all positive but still not physical. The physical solution takes the form of [A B iA+jC]. So C is actually somewhat the difference between A and the 'actual' C.
Is it possible to obtain the physical solution by mixing the left vectors ?
linear-algebra matrices matrix-decomposition svd
asked Aug 14 at 9:49
7E10FC9A
11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For a physical process involving the change in populations of multiple spectral species, one can obtained a bi-linearly degenerate matrix.
It can be decomposed as the outer product of spectral species (column vectors) and kinetics (row vectors), where the number of vectors is the rank of the matrix.
It is known that the column vectors are positive, and it is physically required that the row vectors to be positive as well (negative population is not possible).
However, when performing SVD, it gives sign-changing vectors for both left and right singular vectors.
When performing NNMF, where the vectors are constrained to be positive and the orthogonality loosely holds. The 'left vectors' result of NNMF looks like the left singular vectors in SVD, but either 'shifted up' or have their negative elements replaced by 0.
Assuming the rank of 3, the column vectors it gives are [A B C],which are all positive but still not physical. The physical solution takes the form of [A B iA+jC]. So C is actually somewhat the difference between A and the 'actual' C.
Is it possible to obtain the physical solution by mixing the left vectors ?
linear-algebra matrices matrix-decomposition svd
asked Aug 14 at 9:49
7E10FC9A
11
For a physical process involving the change in populations of multiple spectral species, one can obtained a bi-linearly degenerate matrix.
It can be decomposed as the outer product of spectral species (column vectors) and kinetics (row vectors), where the number of vectors is the rank of the matrix.
It is known that the column vectors are positive, and it is physically required that the row vectors to be positive as well (negative population is not possible).
However, when performing SVD, it gives sign-changing vectors for both left and right singular vectors.
When performing NNMF, where the vectors are constrained to be positive and the orthogonality loosely holds. The 'left vectors' result of NNMF looks like the left singular vectors in SVD, but either 'shifted up' or have their negative elements replaced by 0.
Assuming the rank of 3, the column vectors it gives are [A B C],which are all positive but still not physical. The physical solution takes the form of [A B iA+jC]. So C is actually somewhat the difference between A and the 'actual' C.
Is it possible to obtain the physical solution by mixing the left vectors ?
linear-algebra matrices matrix-decomposition svd
asked Aug 14 at 9:49
7E10FC9A
11
asked Aug 14 at 9:49
7E10FC9A
11
asked Aug 14 at 9:49
7E10FC9A
11
asked Aug 14 at 9:49
7E10FC9A
11
11
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add a comment |Â
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