Understanding about $S^n_+=cap_{z in R^n}S_z$,so it is convex.
A set $S^{+}$ of positive semi-definite matrices (PSD) is defined as
$S^+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n,mathbf X^T=X }$
Use the property of which intersection of the halfspaces is also convex to prove
$S^+$ is convex set.
And the solution is as below
Define $S_z={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0}={mathbf X in S^n |Tr(mathbf X mathbf zmathbf z^T )ge 0}={mathbf X in S^n |Tr(mathbf X mathbf Z)ge 0}$,so $S_z$ is a halfspace,so $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,so it is convex.
I can't understand the last part of the solution,i mean the relation between
1.$S_z$ is a halfspace
2.$S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$
3.so it is convex.
By the way,why is $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,i can't understand this either,can anyone explain them to me?
convex-analysis convex-optimization
asked Dec 2 at 2:56
electronic component
387
add a comment |
A set $S^{+}$ of positive semi-definite matrices (PSD) is defined as
$S^+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n,mathbf X^T=X }$
Use the property of which intersection of the halfspaces is also convex to prove
$S^+$ is convex set.
And the solution is as below
Define $S_z={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0}={mathbf X in S^n |Tr(mathbf X mathbf zmathbf z^T )ge 0}={mathbf X in S^n |Tr(mathbf X mathbf Z)ge 0}$,so $S_z$ is a halfspace,so $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,so it is convex.
I can't understand the last part of the solution,i mean the relation between
1.$S_z$ is a halfspace
2.$S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$
3.so it is convex.
By the way,why is $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,i can't understand this either,can anyone explain them to me?
convex-analysis convex-optimization
asked Dec 2 at 2:56
electronic component
387
add a comment |
A set $S^{+}$ of positive semi-definite matrices (PSD) is defined as
$S^+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n,mathbf X^T=X }$
Use the property of which intersection of the halfspaces is also convex to prove
$S^+$ is convex set.
And the solution is as below
Define $S_z={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0}={mathbf X in S^n |Tr(mathbf X mathbf zmathbf z^T )ge 0}={mathbf X in S^n |Tr(mathbf X mathbf Z)ge 0}$,so $S_z$ is a halfspace,so $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,so it is convex.
I can't understand the last part of the solution,i mean the relation between
1.$S_z$ is a halfspace
2.$S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$
3.so it is convex.
By the way,why is $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,i can't understand this either,can anyone explain them to me?
convex-analysis convex-optimization
asked Dec 2 at 2:56
electronic component
387
A set $S^{+}$ of positive semi-definite matrices (PSD) is defined as
$S^+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n,mathbf X^T=X }$
Use the property of which intersection of the halfspaces is also convex to prove
$S^+$ is convex set.
And the solution is as below
Define $S_z={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0}={mathbf X in S^n |Tr(mathbf X mathbf zmathbf z^T )ge 0}={mathbf X in S^n |Tr(mathbf X mathbf Z)ge 0}$,so $S_z$ is a halfspace,so $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,so it is convex.
I can't understand the last part of the solution,i mean the relation between
1.$S_z$ is a halfspace
2.$S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$
3.so it is convex.
By the way,why is $S^n_+={mathbf X in S^n |mathbf z^T mathbf X mathbf z ge 0,forall z in R^n }=cap_{z in R^n}S_z$,i can't understand this either,can anyone explain them to me?
convex-analysis convex-optimization
convex-analysis convex-optimization
asked Dec 2 at 2:56
electronic component
387
asked Dec 2 at 2:56
electronic component
387
asked Dec 2 at 2:56
electronic component
387
asked Dec 2 at 2:56
electronic component
387
asked Dec 2 at 2:56
electronic component
387
387
add a comment |
add a comment |
1 Answer
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The following steps are equivalent:
$X$ is positive-semidefinite.- By definition, $forall zin{Bbb R}^n$ we have $z^TXzge 0$ (that is, $Xin S_+^n$).
$forall zin{Bbb R}^n$ we have $z^TXzge 0$ can be interpreted as $forall zin{Bbb R}^n$ it holds $Xin S_z$.- By definition of intersection, the latter gives $Xin cap_{zin{Bbb R}^n}S_z$.
In particular, equivalence of $2$ and $4$ means that $S_+^n=cap_{zin{Bbb R}^n}S_z$. $S_z$ is a half-space as it is given by inequality "the inner product with a fixed vector is non-negative". A half-space is convex, hence, $S_+^n$ is convex as an intersection of convex sets.
answered Dec 2 at 17:10
A.Γ.
21.8k22455
add a comment |
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The following steps are equivalent:
$X$ is positive-semidefinite.- By definition, $forall zin{Bbb R}^n$ we have $z^TXzge 0$ (that is, $Xin S_+^n$).
$forall zin{Bbb R}^n$ we have $z^TXzge 0$ can be interpreted as $forall zin{Bbb R}^n$ it holds $Xin S_z$.- By definition of intersection, the latter gives $Xin cap_{zin{Bbb R}^n}S_z$.
In particular, equivalence of $2$ and $4$ means that $S_+^n=cap_{zin{Bbb R}^n}S_z$. $S_z$ is a half-space as it is given by inequality "the inner product with a fixed vector is non-negative". A half-space is convex, hence, $S_+^n$ is convex as an intersection of convex sets.
answered Dec 2 at 17:10
A.Γ.
21.8k22455
add a comment |
The following steps are equivalent:
$X$ is positive-semidefinite.- By definition, $forall zin{Bbb R}^n$ we have $z^TXzge 0$ (that is, $Xin S_+^n$).
$forall zin{Bbb R}^n$ we have $z^TXzge 0$ can be interpreted as $forall zin{Bbb R}^n$ it holds $Xin S_z$.- By definition of intersection, the latter gives $Xin cap_{zin{Bbb R}^n}S_z$.
In particular, equivalence of $2$ and $4$ means that $S_+^n=cap_{zin{Bbb R}^n}S_z$. $S_z$ is a half-space as it is given by inequality "the inner product with a fixed vector is non-negative". A half-space is convex, hence, $S_+^n$ is convex as an intersection of convex sets.
answered Dec 2 at 17:10
A.Γ.
21.8k22455
add a comment |
The following steps are equivalent:
$X$ is positive-semidefinite.- By definition, $forall zin{Bbb R}^n$ we have $z^TXzge 0$ (that is, $Xin S_+^n$).
$forall zin{Bbb R}^n$ we have $z^TXzge 0$ can be interpreted as $forall zin{Bbb R}^n$ it holds $Xin S_z$.- By definition of intersection, the latter gives $Xin cap_{zin{Bbb R}^n}S_z$.
In particular, equivalence of $2$ and $4$ means that $S_+^n=cap_{zin{Bbb R}^n}S_z$. $S_z$ is a half-space as it is given by inequality "the inner product with a fixed vector is non-negative". A half-space is convex, hence, $S_+^n$ is convex as an intersection of convex sets.
answered Dec 2 at 17:10
A.Γ.
21.8k22455
The following steps are equivalent:
$X$ is positive-semidefinite.- By definition, $forall zin{Bbb R}^n$ we have $z^TXzge 0$ (that is, $Xin S_+^n$).
$forall zin{Bbb R}^n$ we have $z^TXzge 0$ can be interpreted as $forall zin{Bbb R}^n$ it holds $Xin S_z$.- By definition of intersection, the latter gives $Xin cap_{zin{Bbb R}^n}S_z$.
In particular, equivalence of $2$ and $4$ means that $S_+^n=cap_{zin{Bbb R}^n}S_z$. $S_z$ is a half-space as it is given by inequality "the inner product with a fixed vector is non-negative". A half-space is convex, hence, $S_+^n$ is convex as an intersection of convex sets.
answered Dec 2 at 17:10
A.Γ.
21.8k22455
answered Dec 2 at 17:10
A.Γ.
21.8k22455
answered Dec 2 at 17:10
A.Γ.
21.8k22455
answered Dec 2 at 17:10
A.Γ.
21.8k22455
21.8k22455
add a comment |
add a comment |
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