What can be the convex relaxation of a quadratic matrix inequality?
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I am trying to relax the Quadratic Matrix Inequality given as:
$$W leq X^TX+Y^TY \ Wgeq 0 $$ Here, $X,Y,W in mathbb{R}^{ntimes n}$ matrices. These two are to be solved along with one linear matrix inequality in $W, X,$ and $Y$.
My attempt:
I tried applying the Schur Complement lemma as:
$$begin{bmatrix} X^TX-W& Y^T\ Y & I end{bmatrix}geq 0$$
But this is making the problem infeasible as I understand that there is no solution such that $X^TXgeq W$ along with other constraints.
how can I relax this Quadratic Matrix Inequality?
convex-optimization nonlinear-optimization semidefinite-programming relaxations
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
$endgroup$
add a comment |
$begingroup$
I am trying to relax the Quadratic Matrix Inequality given as:
$$W leq X^TX+Y^TY \ Wgeq 0 $$ Here, $X,Y,W in mathbb{R}^{ntimes n}$ matrices. These two are to be solved along with one linear matrix inequality in $W, X,$ and $Y$.
My attempt:
I tried applying the Schur Complement lemma as:
$$begin{bmatrix} X^TX-W& Y^T\ Y & I end{bmatrix}geq 0$$
But this is making the problem infeasible as I understand that there is no solution such that $X^TXgeq W$ along with other constraints.
how can I relax this Quadratic Matrix Inequality?
convex-optimization nonlinear-optimization semidefinite-programming relaxations
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
$endgroup$
$begingroup$
the squared term is on the wrong side of the inequality for the Schur complement; perhaps you can substitute $Z=X^TX$, $Zsucceq 0$.
$endgroup$
– LinAlg
Dec 6 '18 at 21:47
add a comment |
$begingroup$
I am trying to relax the Quadratic Matrix Inequality given as:
$$W leq X^TX+Y^TY \ Wgeq 0 $$ Here, $X,Y,W in mathbb{R}^{ntimes n}$ matrices. These two are to be solved along with one linear matrix inequality in $W, X,$ and $Y$.
My attempt:
I tried applying the Schur Complement lemma as:
$$begin{bmatrix} X^TX-W& Y^T\ Y & I end{bmatrix}geq 0$$
But this is making the problem infeasible as I understand that there is no solution such that $X^TXgeq W$ along with other constraints.
how can I relax this Quadratic Matrix Inequality?
convex-optimization nonlinear-optimization semidefinite-programming relaxations
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
$endgroup$
I am trying to relax the Quadratic Matrix Inequality given as:
$$W leq X^TX+Y^TY \ Wgeq 0 $$ Here, $X,Y,W in mathbb{R}^{ntimes n}$ matrices. These two are to be solved along with one linear matrix inequality in $W, X,$ and $Y$.
My attempt:
I tried applying the Schur Complement lemma as:
$$begin{bmatrix} X^TX-W& Y^T\ Y & I end{bmatrix}geq 0$$
But this is making the problem infeasible as I understand that there is no solution such that $X^TXgeq W$ along with other constraints.
how can I relax this Quadratic Matrix Inequality?
convex-optimization nonlinear-optimization semidefinite-programming relaxations
convex-optimization nonlinear-optimization semidefinite-programming relaxations
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
asked Dec 6 '18 at 7:47
Parikshit PareekParikshit Pareek
336
336
$begingroup$
the squared term is on the wrong side of the inequality for the Schur complement; perhaps you can substitute $Z=X^TX$, $Zsucceq 0$.
$endgroup$
– LinAlg
Dec 6 '18 at 21:47
add a comment |
$begingroup$
the squared term is on the wrong side of the inequality for the Schur complement; perhaps you can substitute $Z=X^TX$, $Zsucceq 0$.
$endgroup$
– LinAlg
Dec 6 '18 at 21:47
$begingroup$
the squared term is on the wrong side of the inequality for the Schur complement; perhaps you can substitute $Z=X^TX$, $Zsucceq 0$.
$endgroup$
– LinAlg
Dec 6 '18 at 21:47
$begingroup$
the squared term is on the wrong side of the inequality for the Schur complement; perhaps you can substitute $Z=X^TX$, $Zsucceq 0$.
$endgroup$
– LinAlg
Dec 6 '18 at 21:47
add a comment |
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$begingroup$
the squared term is on the wrong side of the inequality for the Schur complement; perhaps you can substitute $Z=X^TX$, $Zsucceq 0$.
$endgroup$
– LinAlg
Dec 6 '18 at 21:47