Rank-one positive decomposition for a entry-wise positive positive definite matrixCondition for doubly non-negative matrices to be completely positiveDecomposition of a semi-definite matrix into sums of sparse semi-definite matricesProof for a Rank-One Decomposition Theorem of Positive (semi) Definite MatricesFull rank submatrices of positive semidefinite matrixPositive semi-definite in the limitOptimization problem with determinant as objectiveIs this graph problem NP-Hard?
Rank-one positive decomposition for a entry-wise positive positive definite matrix
Condition for doubly non-negative matrices to be completely positiveDecomposition of a semi-definite matrix into sums of sparse semi-definite matricesProof for a Rank-One Decomposition Theorem of Positive (semi) Definite MatricesFull rank submatrices of positive semidefinite matrixPositive semi-definite in the limitOptimization problem with determinant as objectiveIs this graph problem NP-Hard?
$begingroup$
I have asked this question in math.se without any success.
Let $mathbfA$ be a symmetric $ntimes n$ positive semi-definite matrix and also such that each of its entries is positive. Does $mathbfA$ have a decomposition of the form
beginalign
mathbfA ,=,sum_i=1^kmathbfy_imathbfy_i^T
endalign
where each vector $mathbfy$ is also entry-wise positive and $kleq n$.
linear-algebra matrices
edited Sep 26 at 6:14
YCor
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asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
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add a comment
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$begingroup$
I have asked this question in math.se without any success.
Let $mathbfA$ be a symmetric $ntimes n$ positive semi-definite matrix and also such that each of its entries is positive. Does $mathbfA$ have a decomposition of the form
beginalign
mathbfA ,=,sum_i=1^kmathbfy_imathbfy_i^T
endalign
where each vector $mathbfy$ is also entry-wise positive and $kleq n$.
linear-algebra matrices
edited Sep 26 at 6:14
YCor
33.1k4 gold badges98 silver badges152 bronze badges
asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
$endgroup$
2
$begingroup$
Your matrix is doubly nonnegative mathworld.wolfram.com/DoublyNonnegativeMatrix.html and your question is equivalent to asking if every such matrix is completely positive. mathworld.wolfram.com/CompletelyPositiveMatrix.html Unfortunaley mathworld only states the other direction which is trivial but maybe you find something in the references.
$endgroup$
– user35593
Sep 26 at 7:00
add a comment
|
$begingroup$
I have asked this question in math.se without any success.
Let $mathbfA$ be a symmetric $ntimes n$ positive semi-definite matrix and also such that each of its entries is positive. Does $mathbfA$ have a decomposition of the form
beginalign
mathbfA ,=,sum_i=1^kmathbfy_imathbfy_i^T
endalign
where each vector $mathbfy$ is also entry-wise positive and $kleq n$.
linear-algebra matrices
edited Sep 26 at 6:14
YCor
33.1k4 gold badges98 silver badges152 bronze badges
asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
$endgroup$
I have asked this question in math.se without any success.
Let $mathbfA$ be a symmetric $ntimes n$ positive semi-definite matrix and also such that each of its entries is positive. Does $mathbfA$ have a decomposition of the form
beginalign
mathbfA ,=,sum_i=1^kmathbfy_imathbfy_i^T
endalign
where each vector $mathbfy$ is also entry-wise positive and $kleq n$.
linear-algebra matrices
linear-algebra matrices
edited Sep 26 at 6:14
YCor
33.1k4 gold badges98 silver badges152 bronze badges
asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
edited Sep 26 at 6:14
YCor
33.1k4 gold badges98 silver badges152 bronze badges
asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
edited Sep 26 at 6:14
YCor
33.1k4 gold badges98 silver badges152 bronze badges
edited Sep 26 at 6:14
YCor
33.1k4 gold badges98 silver badges152 bronze badges
edited Sep 26 at 6:14
YCor
33.1k4 gold badges98 silver badges152 bronze badges
33.1k4 gold badges98 silver badges152 bronze badges
asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
asked Sep 26 at 4:04
dineshdileepdineshdileep
1,2459 silver badges12 bronze badges
1,2459 silver badges12 bronze badges
2
$begingroup$
Your matrix is doubly nonnegative mathworld.wolfram.com/DoublyNonnegativeMatrix.html and your question is equivalent to asking if every such matrix is completely positive. mathworld.wolfram.com/CompletelyPositiveMatrix.html Unfortunaley mathworld only states the other direction which is trivial but maybe you find something in the references.
$endgroup$
– user35593
Sep 26 at 7:00
add a comment
|
2
$begingroup$
Your matrix is doubly nonnegative mathworld.wolfram.com/DoublyNonnegativeMatrix.html and your question is equivalent to asking if every such matrix is completely positive. mathworld.wolfram.com/CompletelyPositiveMatrix.html Unfortunaley mathworld only states the other direction which is trivial but maybe you find something in the references.
$endgroup$
– user35593
Sep 26 at 7:00
2
2
$begingroup$
Your matrix is doubly nonnegative mathworld.wolfram.com/DoublyNonnegativeMatrix.html and your question is equivalent to asking if every such matrix is completely positive. mathworld.wolfram.com/CompletelyPositiveMatrix.html Unfortunaley mathworld only states the other direction which is trivial but maybe you find something in the references.
$endgroup$
– user35593
Sep 26 at 7:00
$begingroup$
Your matrix is doubly nonnegative mathworld.wolfram.com/DoublyNonnegativeMatrix.html and your question is equivalent to asking if every such matrix is completely positive. mathworld.wolfram.com/CompletelyPositiveMatrix.html Unfortunaley mathworld only states the other direction which is trivial but maybe you find something in the references.
$endgroup$
– user35593
Sep 26 at 7:00
add a comment
|
2 Answers
2
active
oldest
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$begingroup$
No, there exist doubly nonnegative matrices which are not completely positive, see for example The difference between 5 x 5 doubly nonnegative and completely positive matrices (2009).
A graph-based characterization of doubly nonnegative matrices which are completely positive is: Every doubly nonnegative matrix realization of a graph $G$ is completely positive if and only if $G$ does not contain an odd cycle of length at least 5, see Open problems in the theory of completely positive and copositive matrices (2015).
An alternative characterization is give in a 2011 MO question: A doubly nonnegative $ntimes n$ matrix is completely positive if and only if the $n$
vectors making up the Gram matrix lie in the non-negative orthant of some space of dimension $>n$.
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
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$begingroup$
Not necessarily.
If this would hold, all components of the $y_i$ would be bounded in terms of $A$; so, by compactness argument, the same would hold for matrices and columns with non-negative entries. We will show that this is not the case.
Let $I$ and $J$ be the identty and the all-ones matrices of order $3times3$. Set
$$
A=left[matrix 100I&J\J&100Iright].
$$
Then each $y_i$ may contain at most one non-zero among the first three entries, the same for the last three. On the other hand, for each $a=1,2,3$ and $b=4,5,6$ there should be a $y_i$ with non-zero $a$th and $b$th entries. Thus $kgeq9$.
Remark. This argument shows that, in general, $kgeq [n^2/4]$. It cannot show more, since the edges of an arbitrary graph on $n$ vertices may be split into at most $n^2/4$ cliques. It may happen that this estimate is sharp.
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
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2 Answers
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active
oldest
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2 Answers
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active
oldest
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votes
$begingroup$
No, there exist doubly nonnegative matrices which are not completely positive, see for example The difference between 5 x 5 doubly nonnegative and completely positive matrices (2009).
A graph-based characterization of doubly nonnegative matrices which are completely positive is: Every doubly nonnegative matrix realization of a graph $G$ is completely positive if and only if $G$ does not contain an odd cycle of length at least 5, see Open problems in the theory of completely positive and copositive matrices (2015).
An alternative characterization is give in a 2011 MO question: A doubly nonnegative $ntimes n$ matrix is completely positive if and only if the $n$
vectors making up the Gram matrix lie in the non-negative orthant of some space of dimension $>n$.
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
$endgroup$
add a comment
|
$begingroup$
No, there exist doubly nonnegative matrices which are not completely positive, see for example The difference between 5 x 5 doubly nonnegative and completely positive matrices (2009).
A graph-based characterization of doubly nonnegative matrices which are completely positive is: Every doubly nonnegative matrix realization of a graph $G$ is completely positive if and only if $G$ does not contain an odd cycle of length at least 5, see Open problems in the theory of completely positive and copositive matrices (2015).
An alternative characterization is give in a 2011 MO question: A doubly nonnegative $ntimes n$ matrix is completely positive if and only if the $n$
vectors making up the Gram matrix lie in the non-negative orthant of some space of dimension $>n$.
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
$endgroup$
add a comment
|
$begingroup$
No, there exist doubly nonnegative matrices which are not completely positive, see for example The difference between 5 x 5 doubly nonnegative and completely positive matrices (2009).
A graph-based characterization of doubly nonnegative matrices which are completely positive is: Every doubly nonnegative matrix realization of a graph $G$ is completely positive if and only if $G$ does not contain an odd cycle of length at least 5, see Open problems in the theory of completely positive and copositive matrices (2015).
An alternative characterization is give in a 2011 MO question: A doubly nonnegative $ntimes n$ matrix is completely positive if and only if the $n$
vectors making up the Gram matrix lie in the non-negative orthant of some space of dimension $>n$.
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
$endgroup$
No, there exist doubly nonnegative matrices which are not completely positive, see for example The difference between 5 x 5 doubly nonnegative and completely positive matrices (2009).
A graph-based characterization of doubly nonnegative matrices which are completely positive is: Every doubly nonnegative matrix realization of a graph $G$ is completely positive if and only if $G$ does not contain an odd cycle of length at least 5, see Open problems in the theory of completely positive and copositive matrices (2015).
An alternative characterization is give in a 2011 MO question: A doubly nonnegative $ntimes n$ matrix is completely positive if and only if the $n$
vectors making up the Gram matrix lie in the non-negative orthant of some space of dimension $>n$.
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
edited Sep 26 at 7:42
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
answered Sep 26 at 7:23
Carlo BeenakkerCarlo Beenakker
92.7k9 gold badges229 silver badges341 bronze badges
92.7k9 gold badges229 silver badges341 bronze badges
add a comment
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add a comment
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$begingroup$
Not necessarily.
If this would hold, all components of the $y_i$ would be bounded in terms of $A$; so, by compactness argument, the same would hold for matrices and columns with non-negative entries. We will show that this is not the case.
Let $I$ and $J$ be the identty and the all-ones matrices of order $3times3$. Set
$$
A=left[matrix 100I&J\J&100Iright].
$$
Then each $y_i$ may contain at most one non-zero among the first three entries, the same for the last three. On the other hand, for each $a=1,2,3$ and $b=4,5,6$ there should be a $y_i$ with non-zero $a$th and $b$th entries. Thus $kgeq9$.
Remark. This argument shows that, in general, $kgeq [n^2/4]$. It cannot show more, since the edges of an arbitrary graph on $n$ vertices may be split into at most $n^2/4$ cliques. It may happen that this estimate is sharp.
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
$endgroup$
add a comment
|
$begingroup$
Not necessarily.
If this would hold, all components of the $y_i$ would be bounded in terms of $A$; so, by compactness argument, the same would hold for matrices and columns with non-negative entries. We will show that this is not the case.
Let $I$ and $J$ be the identty and the all-ones matrices of order $3times3$. Set
$$
A=left[matrix 100I&J\J&100Iright].
$$
Then each $y_i$ may contain at most one non-zero among the first three entries, the same for the last three. On the other hand, for each $a=1,2,3$ and $b=4,5,6$ there should be a $y_i$ with non-zero $a$th and $b$th entries. Thus $kgeq9$.
Remark. This argument shows that, in general, $kgeq [n^2/4]$. It cannot show more, since the edges of an arbitrary graph on $n$ vertices may be split into at most $n^2/4$ cliques. It may happen that this estimate is sharp.
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
$endgroup$
add a comment
|
$begingroup$
Not necessarily.
If this would hold, all components of the $y_i$ would be bounded in terms of $A$; so, by compactness argument, the same would hold for matrices and columns with non-negative entries. We will show that this is not the case.
Let $I$ and $J$ be the identty and the all-ones matrices of order $3times3$. Set
$$
A=left[matrix 100I&J\J&100Iright].
$$
Then each $y_i$ may contain at most one non-zero among the first three entries, the same for the last three. On the other hand, for each $a=1,2,3$ and $b=4,5,6$ there should be a $y_i$ with non-zero $a$th and $b$th entries. Thus $kgeq9$.
Remark. This argument shows that, in general, $kgeq [n^2/4]$. It cannot show more, since the edges of an arbitrary graph on $n$ vertices may be split into at most $n^2/4$ cliques. It may happen that this estimate is sharp.
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
$endgroup$
Not necessarily.
If this would hold, all components of the $y_i$ would be bounded in terms of $A$; so, by compactness argument, the same would hold for matrices and columns with non-negative entries. We will show that this is not the case.
Let $I$ and $J$ be the identty and the all-ones matrices of order $3times3$. Set
$$
A=left[matrix 100I&J\J&100Iright].
$$
Then each $y_i$ may contain at most one non-zero among the first three entries, the same for the last three. On the other hand, for each $a=1,2,3$ and $b=4,5,6$ there should be a $y_i$ with non-zero $a$th and $b$th entries. Thus $kgeq9$.
Remark. This argument shows that, in general, $kgeq [n^2/4]$. It cannot show more, since the edges of an arbitrary graph on $n$ vertices may be split into at most $n^2/4$ cliques. It may happen that this estimate is sharp.
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
answered Sep 26 at 7:23
Ilya BogdanovIlya Bogdanov
15.6k30 silver badges66 bronze badges
15.6k30 silver badges66 bronze badges
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$begingroup$
Your matrix is doubly nonnegative mathworld.wolfram.com/DoublyNonnegativeMatrix.html and your question is equivalent to asking if every such matrix is completely positive. mathworld.wolfram.com/CompletelyPositiveMatrix.html Unfortunaley mathworld only states the other direction which is trivial but maybe you find something in the references.
$endgroup$
– user35593
Sep 26 at 7:00