Why can $X=qq^T in S^n_+$ prove $Y notin mathbf K^*$?
$begingroup$
$mathbf K = S^n_+$ (nxn real symmetric positive semidefinite matrix). Show that the
dual of $mathbf K$
$mathbf K^*={mathbf Y | Tr(mathbf X mathbf Y) >0, forall mathbf X ge mathbf 0}$
is self-dual, i.e., $mathbf K^* = mathbf K$.
Hint :Prove $mathbf K^* subseteq S^n_+$ and $S^n_+ subseteq mathbf K^*$
Here is parts of its solution
$1.$Prove $mathbf K^* subseteq S^n_+$
let $Y in mathbf K^*$,we need to prove $Y in S^n_+ $
$1-1.$ Suppose $ Y notin S^n_+$,there exists $q in R^n$ so that $q^TYq=Tr(Yqq^T) le 0$
$1-2.$ Let $X=qq^T in S^n_+$,then $tr(XY) lt 0$,it means that $Y notin mathbf K^*$ (Contradiction) ,so $Y in S^n_+$
I have three problem in here
1.Why should we let $mathbf X$ is $qq^T in S^n_+$
2.Why is $X=qq^T in S^n_+$,then $tr(XY) lt 0$,shouldn't it be $tr(XY) gt 0$ ?
3.Why can $X=qq^T in S^n_+$ prove $Y notin mathbf K^*$?
4.Why can $Y notin mathbf K^*$ prove $mathbf K^* in S_+^n$ ?
Can anyone explain it to me?
convex-analysis convex-optimization
asked Dec 5 '18 at 23:25
shineeleshineele
377
$endgroup$
add a comment |
$begingroup$
$mathbf K = S^n_+$ (nxn real symmetric positive semidefinite matrix). Show that the
dual of $mathbf K$
$mathbf K^*={mathbf Y | Tr(mathbf X mathbf Y) >0, forall mathbf X ge mathbf 0}$
is self-dual, i.e., $mathbf K^* = mathbf K$.
Hint :Prove $mathbf K^* subseteq S^n_+$ and $S^n_+ subseteq mathbf K^*$
Here is parts of its solution
$1.$Prove $mathbf K^* subseteq S^n_+$
let $Y in mathbf K^*$,we need to prove $Y in S^n_+ $
$1-1.$ Suppose $ Y notin S^n_+$,there exists $q in R^n$ so that $q^TYq=Tr(Yqq^T) le 0$
$1-2.$ Let $X=qq^T in S^n_+$,then $tr(XY) lt 0$,it means that $Y notin mathbf K^*$ (Contradiction) ,so $Y in S^n_+$
I have three problem in here
1.Why should we let $mathbf X$ is $qq^T in S^n_+$
2.Why is $X=qq^T in S^n_+$,then $tr(XY) lt 0$,shouldn't it be $tr(XY) gt 0$ ?
3.Why can $X=qq^T in S^n_+$ prove $Y notin mathbf K^*$?
4.Why can $Y notin mathbf K^*$ prove $mathbf K^* in S_+^n$ ?
Can anyone explain it to me?
convex-analysis convex-optimization
asked Dec 5 '18 at 23:25
shineeleshineele
377
$endgroup$
add a comment |
$begingroup$
$mathbf K = S^n_+$ (nxn real symmetric positive semidefinite matrix). Show that the
dual of $mathbf K$
$mathbf K^*={mathbf Y | Tr(mathbf X mathbf Y) >0, forall mathbf X ge mathbf 0}$
is self-dual, i.e., $mathbf K^* = mathbf K$.
Hint :Prove $mathbf K^* subseteq S^n_+$ and $S^n_+ subseteq mathbf K^*$
Here is parts of its solution
$1.$Prove $mathbf K^* subseteq S^n_+$
let $Y in mathbf K^*$,we need to prove $Y in S^n_+ $
$1-1.$ Suppose $ Y notin S^n_+$,there exists $q in R^n$ so that $q^TYq=Tr(Yqq^T) le 0$
$1-2.$ Let $X=qq^T in S^n_+$,then $tr(XY) lt 0$,it means that $Y notin mathbf K^*$ (Contradiction) ,so $Y in S^n_+$
I have three problem in here
1.Why should we let $mathbf X$ is $qq^T in S^n_+$
2.Why is $X=qq^T in S^n_+$,then $tr(XY) lt 0$,shouldn't it be $tr(XY) gt 0$ ?
3.Why can $X=qq^T in S^n_+$ prove $Y notin mathbf K^*$?
4.Why can $Y notin mathbf K^*$ prove $mathbf K^* in S_+^n$ ?
Can anyone explain it to me?
convex-analysis convex-optimization
asked Dec 5 '18 at 23:25
shineeleshineele
377
$endgroup$
$mathbf K = S^n_+$ (nxn real symmetric positive semidefinite matrix). Show that the
dual of $mathbf K$
$mathbf K^*={mathbf Y | Tr(mathbf X mathbf Y) >0, forall mathbf X ge mathbf 0}$
is self-dual, i.e., $mathbf K^* = mathbf K$.
Hint :Prove $mathbf K^* subseteq S^n_+$ and $S^n_+ subseteq mathbf K^*$
Here is parts of its solution
$1.$Prove $mathbf K^* subseteq S^n_+$
let $Y in mathbf K^*$,we need to prove $Y in S^n_+ $
$1-1.$ Suppose $ Y notin S^n_+$,there exists $q in R^n$ so that $q^TYq=Tr(Yqq^T) le 0$
$1-2.$ Let $X=qq^T in S^n_+$,then $tr(XY) lt 0$,it means that $Y notin mathbf K^*$ (Contradiction) ,so $Y in S^n_+$
I have three problem in here
1.Why should we let $mathbf X$ is $qq^T in S^n_+$
2.Why is $X=qq^T in S^n_+$,then $tr(XY) lt 0$,shouldn't it be $tr(XY) gt 0$ ?
3.Why can $X=qq^T in S^n_+$ prove $Y notin mathbf K^*$?
4.Why can $Y notin mathbf K^*$ prove $mathbf K^* in S_+^n$ ?
Can anyone explain it to me?
convex-analysis convex-optimization
convex-analysis convex-optimization
asked Dec 5 '18 at 23:25
shineeleshineele
377
asked Dec 5 '18 at 23:25
shineeleshineele
377
asked Dec 5 '18 at 23:25
shineeleshineele
377
asked Dec 5 '18 at 23:25
shineeleshineele
377
asked Dec 5 '18 at 23:25
shineeleshineele
377
377
add a comment |
add a comment |
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