Decomposition of the Outer Product
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I have an $m times n$ matrix that is statistical data. I have a theoretical model about how that data is constructed in which the matrix is the outer product of two vectors. Let's call those vectors $u$ and $v$. Their size is $m times 1$ and $ntimes 1$ respectively such that the outer product is $uv^{ prime}$. The hitch is that there is also missing data, i.e. not all cells in the matrix are in the data. In other words, it is an outer product that is then multiplied cell by cell by a matrix of the same size with zeros and ones. Any idea how I can estimate the vectors $hat{u}$ and $hat{v}$ that yield an outer product matrix that is the closest to the one in the data? I do know where the missing data is (i.e. I know the matrix of zeros and ones that is multiplied by the outer product matrix). Thank you!
matrix-decomposition outer-product
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
$endgroup$
add a comment |
$begingroup$
I have an $m times n$ matrix that is statistical data. I have a theoretical model about how that data is constructed in which the matrix is the outer product of two vectors. Let's call those vectors $u$ and $v$. Their size is $m times 1$ and $ntimes 1$ respectively such that the outer product is $uv^{ prime}$. The hitch is that there is also missing data, i.e. not all cells in the matrix are in the data. In other words, it is an outer product that is then multiplied cell by cell by a matrix of the same size with zeros and ones. Any idea how I can estimate the vectors $hat{u}$ and $hat{v}$ that yield an outer product matrix that is the closest to the one in the data? I do know where the missing data is (i.e. I know the matrix of zeros and ones that is multiplied by the outer product matrix). Thank you!
matrix-decomposition outer-product
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
$endgroup$
$begingroup$
HINT: The columns of the matrix are all scalar multiples of $u$ and the rows are all scalar multiples of $v'$. You won’t be able to recover $u$ and $v$, however, since for any nonzero scalar $k$, $uv'=(ku)(1/k,v')$.
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– amd
Dec 1 '18 at 8:30
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So I could recover them up to a constant k?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:26
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It seems that singular value decomposition (SVD) could help me here. Assuming I fill in the missing values somehow (with weighted row or column averages), I could use SVD to get $M=sum_{i=1}^{k}sigma_i u_i v_{i}'$. This would give me a number of vectors $u$ and $v$. I presume there would be a way to solve for an optimal unique pair that would give the matrix that is the closest to the one from the data. Would the weighting $sigma_i$ help here?
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– Frederic Tremblay
Dec 1 '18 at 20:32
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Also, all values in the matrix of data are between 0 and 1, and all values in $u$ and $v$ should be between 0 and 1.
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 21:17
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Going to the SVD is overkill. Pick any complete nonzero column (which you might’ve had to fill in) and call that $u$. You can then compute the corresponding $v$ with simple division.
$endgroup$
– amd
Dec 2 '18 at 0:08
add a comment |
$begingroup$
I have an $m times n$ matrix that is statistical data. I have a theoretical model about how that data is constructed in which the matrix is the outer product of two vectors. Let's call those vectors $u$ and $v$. Their size is $m times 1$ and $ntimes 1$ respectively such that the outer product is $uv^{ prime}$. The hitch is that there is also missing data, i.e. not all cells in the matrix are in the data. In other words, it is an outer product that is then multiplied cell by cell by a matrix of the same size with zeros and ones. Any idea how I can estimate the vectors $hat{u}$ and $hat{v}$ that yield an outer product matrix that is the closest to the one in the data? I do know where the missing data is (i.e. I know the matrix of zeros and ones that is multiplied by the outer product matrix). Thank you!
matrix-decomposition outer-product
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
$endgroup$
I have an $m times n$ matrix that is statistical data. I have a theoretical model about how that data is constructed in which the matrix is the outer product of two vectors. Let's call those vectors $u$ and $v$. Their size is $m times 1$ and $ntimes 1$ respectively such that the outer product is $uv^{ prime}$. The hitch is that there is also missing data, i.e. not all cells in the matrix are in the data. In other words, it is an outer product that is then multiplied cell by cell by a matrix of the same size with zeros and ones. Any idea how I can estimate the vectors $hat{u}$ and $hat{v}$ that yield an outer product matrix that is the closest to the one in the data? I do know where the missing data is (i.e. I know the matrix of zeros and ones that is multiplied by the outer product matrix). Thank you!
matrix-decomposition outer-product
matrix-decomposition outer-product
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
edited Dec 1 '18 at 8:51
Daniele Tampieri
2,2422722
2,2422722
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
asked Dec 1 '18 at 8:11
Frederic TremblayFrederic Tremblay
1
1
$begingroup$
HINT: The columns of the matrix are all scalar multiples of $u$ and the rows are all scalar multiples of $v'$. You won’t be able to recover $u$ and $v$, however, since for any nonzero scalar $k$, $uv'=(ku)(1/k,v')$.
$endgroup$
– amd
Dec 1 '18 at 8:30
$begingroup$
So I could recover them up to a constant k?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:26
$begingroup$
It seems that singular value decomposition (SVD) could help me here. Assuming I fill in the missing values somehow (with weighted row or column averages), I could use SVD to get $M=sum_{i=1}^{k}sigma_i u_i v_{i}'$. This would give me a number of vectors $u$ and $v$. I presume there would be a way to solve for an optimal unique pair that would give the matrix that is the closest to the one from the data. Would the weighting $sigma_i$ help here?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:32
$begingroup$
Also, all values in the matrix of data are between 0 and 1, and all values in $u$ and $v$ should be between 0 and 1.
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 21:17
$begingroup$
Going to the SVD is overkill. Pick any complete nonzero column (which you might’ve had to fill in) and call that $u$. You can then compute the corresponding $v$ with simple division.
$endgroup$
– amd
Dec 2 '18 at 0:08
add a comment |
$begingroup$
HINT: The columns of the matrix are all scalar multiples of $u$ and the rows are all scalar multiples of $v'$. You won’t be able to recover $u$ and $v$, however, since for any nonzero scalar $k$, $uv'=(ku)(1/k,v')$.
$endgroup$
– amd
Dec 1 '18 at 8:30
$begingroup$
So I could recover them up to a constant k?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:26
$begingroup$
It seems that singular value decomposition (SVD) could help me here. Assuming I fill in the missing values somehow (with weighted row or column averages), I could use SVD to get $M=sum_{i=1}^{k}sigma_i u_i v_{i}'$. This would give me a number of vectors $u$ and $v$. I presume there would be a way to solve for an optimal unique pair that would give the matrix that is the closest to the one from the data. Would the weighting $sigma_i$ help here?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:32
$begingroup$
Also, all values in the matrix of data are between 0 and 1, and all values in $u$ and $v$ should be between 0 and 1.
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 21:17
$begingroup$
Going to the SVD is overkill. Pick any complete nonzero column (which you might’ve had to fill in) and call that $u$. You can then compute the corresponding $v$ with simple division.
$endgroup$
– amd
Dec 2 '18 at 0:08
$begingroup$
HINT: The columns of the matrix are all scalar multiples of $u$ and the rows are all scalar multiples of $v'$. You won’t be able to recover $u$ and $v$, however, since for any nonzero scalar $k$, $uv'=(ku)(1/k,v')$.
$endgroup$
– amd
Dec 1 '18 at 8:30
$begingroup$
HINT: The columns of the matrix are all scalar multiples of $u$ and the rows are all scalar multiples of $v'$. You won’t be able to recover $u$ and $v$, however, since for any nonzero scalar $k$, $uv'=(ku)(1/k,v')$.
$endgroup$
– amd
Dec 1 '18 at 8:30
$begingroup$
So I could recover them up to a constant k?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:26
$begingroup$
So I could recover them up to a constant k?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:26
$begingroup$
It seems that singular value decomposition (SVD) could help me here. Assuming I fill in the missing values somehow (with weighted row or column averages), I could use SVD to get $M=sum_{i=1}^{k}sigma_i u_i v_{i}'$. This would give me a number of vectors $u$ and $v$. I presume there would be a way to solve for an optimal unique pair that would give the matrix that is the closest to the one from the data. Would the weighting $sigma_i$ help here?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:32
$begingroup$
It seems that singular value decomposition (SVD) could help me here. Assuming I fill in the missing values somehow (with weighted row or column averages), I could use SVD to get $M=sum_{i=1}^{k}sigma_i u_i v_{i}'$. This would give me a number of vectors $u$ and $v$. I presume there would be a way to solve for an optimal unique pair that would give the matrix that is the closest to the one from the data. Would the weighting $sigma_i$ help here?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:32
$begingroup$
Also, all values in the matrix of data are between 0 and 1, and all values in $u$ and $v$ should be between 0 and 1.
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 21:17
$begingroup$
Also, all values in the matrix of data are between 0 and 1, and all values in $u$ and $v$ should be between 0 and 1.
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 21:17
$begingroup$
Going to the SVD is overkill. Pick any complete nonzero column (which you might’ve had to fill in) and call that $u$. You can then compute the corresponding $v$ with simple division.
$endgroup$
– amd
Dec 2 '18 at 0:08
$begingroup$
Going to the SVD is overkill. Pick any complete nonzero column (which you might’ve had to fill in) and call that $u$. You can then compute the corresponding $v$ with simple division.
$endgroup$
– amd
Dec 2 '18 at 0:08
add a comment |
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$begingroup$
HINT: The columns of the matrix are all scalar multiples of $u$ and the rows are all scalar multiples of $v'$. You won’t be able to recover $u$ and $v$, however, since for any nonzero scalar $k$, $uv'=(ku)(1/k,v')$.
$endgroup$
– amd
Dec 1 '18 at 8:30
$begingroup$
So I could recover them up to a constant k?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:26
$begingroup$
It seems that singular value decomposition (SVD) could help me here. Assuming I fill in the missing values somehow (with weighted row or column averages), I could use SVD to get $M=sum_{i=1}^{k}sigma_i u_i v_{i}'$. This would give me a number of vectors $u$ and $v$. I presume there would be a way to solve for an optimal unique pair that would give the matrix that is the closest to the one from the data. Would the weighting $sigma_i$ help here?
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 20:32
$begingroup$
Also, all values in the matrix of data are between 0 and 1, and all values in $u$ and $v$ should be between 0 and 1.
$endgroup$
– Frederic Tremblay
Dec 1 '18 at 21:17
$begingroup$
Going to the SVD is overkill. Pick any complete nonzero column (which you might’ve had to fill in) and call that $u$. You can then compute the corresponding $v$ with simple division.
$endgroup$
– amd
Dec 2 '18 at 0:08