Whether the set of a symmetric copositive matrices $x^T A x geq 0$ for $forall x in mathbbR^n$ $x geq 0$ is convex? The Next CEO of Stack OverflowNotation for the set of symmetric matrices and symmetric positive definite matricesIs this set of matrices convex?How to prove that the set of real $ n times n $ PSD matrices of rank $leq r < n $ and unit trace, is not contractible?Convex hull of rotation matrices is closed and contains the originIs the the set of symmetric positive-definite matrices in the space of all symmetric matrices strictly convex?Given $P$ a real symmetric positive semidefinite matrix, what matrix $A$ ensure that $A^TPA$ is PSD?Use the definition of convexity to show that $A$ is a convex set.Show that a set of positive semidefinite (PSD) matrices is a convex setProjection of a Symmetric Matrix onto the space of Positive Semi-Definite MatricesHow can I prove that $mathbbS_+^n$ is a closed and convex set?
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Whether the set of a symmetric copositive matrices $x^T A x geq 0$ for $forall x in mathbbR^n$ $x geq 0$ is convex?
The Next CEO of Stack OverflowNotation for the set of symmetric matrices and symmetric positive definite matricesIs this set of matrices convex?How to prove that the set of real $ n times n $ PSD matrices of rank $leq r < n $ and unit trace, is not contractible?Convex hull of rotation matrices is closed and contains the originIs the the set of symmetric positive-definite matrices in the space of all symmetric matrices strictly convex?Given $P$ a real symmetric positive semidefinite matrix, what matrix $A$ ensure that $A^TPA$ is PSD?Use the definition of convexity to show that $A$ is a convex set.Show that a set of positive semidefinite (PSD) matrices is a convex setProjection of a Symmetric Matrix onto the space of Positive Semi-Definite MatricesHow can I prove that $mathbbS_+^n$ is a closed and convex set?
$begingroup$
Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?
Attempt:
If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?
matrices convex-analysis copositivity
$endgroup$
add a comment |
$begingroup$
Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?
Attempt:
If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?
matrices convex-analysis copositivity
$endgroup$
$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday
$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday
add a comment |
$begingroup$
Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?
Attempt:
If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?
matrices convex-analysis copositivity
$endgroup$
Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?
Attempt:
If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?
matrices convex-analysis copositivity
matrices convex-analysis copositivity
edited yesterday
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked yesterday
learninglearning
617
617
$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday
$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday
add a comment |
$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday
$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday
$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday
$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday
$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday
$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.
$endgroup$
add a comment |
$begingroup$
You can follow these steps.
Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.
Now prove the following two assertions: The first one is,
For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.
Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.
The second is,
For any $A,BinmathcalF$, $A+BinmathcalF$.
Which you can do by calculation $x^T (A+B)x$.
Once done. You can combine those two to show that
For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$
which means, by definition, that $mathcalF$ is convex.
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.
$endgroup$
add a comment |
$begingroup$
No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.
$endgroup$
add a comment |
$begingroup$
No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.
$endgroup$
No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
$begingroup$
You can follow these steps.
Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.
Now prove the following two assertions: The first one is,
For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.
Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.
The second is,
For any $A,BinmathcalF$, $A+BinmathcalF$.
Which you can do by calculation $x^T (A+B)x$.
Once done. You can combine those two to show that
For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$
which means, by definition, that $mathcalF$ is convex.
$endgroup$
add a comment |
$begingroup$
You can follow these steps.
Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.
Now prove the following two assertions: The first one is,
For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.
Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.
The second is,
For any $A,BinmathcalF$, $A+BinmathcalF$.
Which you can do by calculation $x^T (A+B)x$.
Once done. You can combine those two to show that
For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$
which means, by definition, that $mathcalF$ is convex.
$endgroup$
add a comment |
$begingroup$
You can follow these steps.
Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.
Now prove the following two assertions: The first one is,
For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.
Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.
The second is,
For any $A,BinmathcalF$, $A+BinmathcalF$.
Which you can do by calculation $x^T (A+B)x$.
Once done. You can combine those two to show that
For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$
which means, by definition, that $mathcalF$ is convex.
$endgroup$
You can follow these steps.
Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.
Now prove the following two assertions: The first one is,
For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.
Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.
The second is,
For any $A,BinmathcalF$, $A+BinmathcalF$.
Which you can do by calculation $x^T (A+B)x$.
Once done. You can combine those two to show that
For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$
which means, by definition, that $mathcalF$ is convex.
answered yesterday
YankoYanko
8,1282830
8,1282830
add a comment |
add a comment |
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$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday
$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday