Adding trace operator to a constraint in optimization? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Constrained optimization: equality constraintOptimization with symmetric matrix constraintRelaxed optimization problemsIntuitive explanation of trace minimization under constraintsAfter solving binary programing problem the constrait is violatedWhy constraint optimization problem becomes infeasible when coefficient is zeroWhat is meant in this optimization problemDerivative-free, Simulation-based OptimizationMinimizing quadratic objective subject to a quadratic equality constraintSolving dynamic optimization with non-binding inequality constraint
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Adding trace operator to a constraint in optimization?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Constrained optimization: equality constraintOptimization with symmetric matrix constraintRelaxed optimization problemsIntuitive explanation of trace minimization under constraintsAfter solving binary programing problem the constrait is violatedWhy constraint optimization problem becomes infeasible when coefficient is zeroWhat is meant in this optimization problemDerivative-free, Simulation-based OptimizationMinimizing quadratic objective subject to a quadratic equality constraintSolving dynamic optimization with non-binding inequality constraint
$begingroup$
MIN tr(ABCD)
s.t BC=I
An optimization problem with trace in the objective function, but matrix equation in the constraint. So, you see the problem right away when constructing the Lagrangian (you end up adding scalar with matrices).
I'm reading the solution and it goes like this when building the Lagrangian:
L=tr(ABCD) - tr((BC-I)*Z)
Can you explain this solution? Where did the Z come from? how does it help solving the problem mentioned above. How was tr() added to the constraint.
linear-algebra matrices optimization
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
MIN tr(ABCD)
s.t BC=I
An optimization problem with trace in the objective function, but matrix equation in the constraint. So, you see the problem right away when constructing the Lagrangian (you end up adding scalar with matrices).
I'm reading the solution and it goes like this when building the Lagrangian:
L=tr(ABCD) - tr((BC-I)*Z)
Can you explain this solution? Where did the Z come from? how does it help solving the problem mentioned above. How was tr() added to the constraint.
linear-algebra matrices optimization
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
PS: that wasn't the exact problem, just a mock, but it conveys the idea.
$endgroup$
– John Deterious
Apr 8 at 19:24
$begingroup$
Yeah, the $Z$ is probably a $2$.
$endgroup$
– Mark Fischler
Apr 8 at 19:24
$begingroup$
@MarkFischler No, it is a symmetric matrix. Somehow it helps.
$endgroup$
– John Deterious
Apr 8 at 19:25
$begingroup$
And the condition that $BC = I$ can't be what you meant, because this becomes finding $min mbox tr(AD)$.
$endgroup$
– Mark Fischler
Apr 8 at 19:26
$begingroup$
@MarkFischler This is just a toy example, you are correct. But, still, regardless of that point, what's the answer?
$endgroup$
– John Deterious
Apr 8 at 19:27
add a comment |
$begingroup$
MIN tr(ABCD)
s.t BC=I
An optimization problem with trace in the objective function, but matrix equation in the constraint. So, you see the problem right away when constructing the Lagrangian (you end up adding scalar with matrices).
I'm reading the solution and it goes like this when building the Lagrangian:
L=tr(ABCD) - tr((BC-I)*Z)
Can you explain this solution? Where did the Z come from? how does it help solving the problem mentioned above. How was tr() added to the constraint.
linear-algebra matrices optimization
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
MIN tr(ABCD)
s.t BC=I
An optimization problem with trace in the objective function, but matrix equation in the constraint. So, you see the problem right away when constructing the Lagrangian (you end up adding scalar with matrices).
I'm reading the solution and it goes like this when building the Lagrangian:
L=tr(ABCD) - tr((BC-I)*Z)
Can you explain this solution? Where did the Z come from? how does it help solving the problem mentioned above. How was tr() added to the constraint.
linear-algebra matrices optimization
linear-algebra matrices optimization
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 19:49
John Deterious
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
asked Apr 8 at 19:23
John DeteriousJohn Deterious
83
83
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
John Deterious is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
PS: that wasn't the exact problem, just a mock, but it conveys the idea.
$endgroup$
– John Deterious
Apr 8 at 19:24
$begingroup$
Yeah, the $Z$ is probably a $2$.
$endgroup$
– Mark Fischler
Apr 8 at 19:24
$begingroup$
@MarkFischler No, it is a symmetric matrix. Somehow it helps.
$endgroup$
– John Deterious
Apr 8 at 19:25
$begingroup$
And the condition that $BC = I$ can't be what you meant, because this becomes finding $min mbox tr(AD)$.
$endgroup$
– Mark Fischler
Apr 8 at 19:26
$begingroup$
@MarkFischler This is just a toy example, you are correct. But, still, regardless of that point, what's the answer?
$endgroup$
– John Deterious
Apr 8 at 19:27
add a comment |
$begingroup$
PS: that wasn't the exact problem, just a mock, but it conveys the idea.
$endgroup$
– John Deterious
Apr 8 at 19:24
$begingroup$
Yeah, the $Z$ is probably a $2$.
$endgroup$
– Mark Fischler
Apr 8 at 19:24
$begingroup$
@MarkFischler No, it is a symmetric matrix. Somehow it helps.
$endgroup$
– John Deterious
Apr 8 at 19:25
$begingroup$
And the condition that $BC = I$ can't be what you meant, because this becomes finding $min mbox tr(AD)$.
$endgroup$
– Mark Fischler
Apr 8 at 19:26
$begingroup$
@MarkFischler This is just a toy example, you are correct. But, still, regardless of that point, what's the answer?
$endgroup$
– John Deterious
Apr 8 at 19:27
$begingroup$
PS: that wasn't the exact problem, just a mock, but it conveys the idea.
$endgroup$
– John Deterious
Apr 8 at 19:24
$begingroup$
PS: that wasn't the exact problem, just a mock, but it conveys the idea.
$endgroup$
– John Deterious
Apr 8 at 19:24
$begingroup$
Yeah, the $Z$ is probably a $2$.
$endgroup$
– Mark Fischler
Apr 8 at 19:24
$begingroup$
Yeah, the $Z$ is probably a $2$.
$endgroup$
– Mark Fischler
Apr 8 at 19:24
$begingroup$
@MarkFischler No, it is a symmetric matrix. Somehow it helps.
$endgroup$
– John Deterious
Apr 8 at 19:25
$begingroup$
@MarkFischler No, it is a symmetric matrix. Somehow it helps.
$endgroup$
– John Deterious
Apr 8 at 19:25
$begingroup$
And the condition that $BC = I$ can't be what you meant, because this becomes finding $min mbox tr(AD)$.
$endgroup$
– Mark Fischler
Apr 8 at 19:26
$begingroup$
And the condition that $BC = I$ can't be what you meant, because this becomes finding $min mbox tr(AD)$.
$endgroup$
– Mark Fischler
Apr 8 at 19:26
$begingroup$
@MarkFischler This is just a toy example, you are correct. But, still, regardless of that point, what's the answer?
$endgroup$
– John Deterious
Apr 8 at 19:27
$begingroup$
@MarkFischler This is just a toy example, you are correct. But, still, regardless of that point, what's the answer?
$endgroup$
– John Deterious
Apr 8 at 19:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think, you need the trace in this setting, since the constraint is not scalar, but matrix-valued. The $Z$ is your matrix-valued Lagrange multiplier, which cannot be scalar in this setting.
I have written down a the Lagrange approach for such very general equality constraints:
beginalign
max_x in X f(x) \ texts.t. g(x) = 0,
endalign
where $f:X rightarrow mathbbR$, and $g: X rightarrow Y$, where $X$ and $Y$ are Hilbert spaces (like $mathbbR^k$, $mathcalL^2$,...).
The Lagrange function is given by
$$L(x, Lambda) = f(x) + langle Lambda, g(x) rangle_Y,$$
where $Lambda in Y$ is the Lagrange multiplier, and $langle cdot, cdot rangle_Y$ is the inner product of $Y$.
The question is: Does the set of matrices that is some superset of the image of $BC-I$ in your notation, form a Hilbert space $Y$ with the trace of the product of two matrices being their inner product; i.e. $langle A, B rangle_X := mathrmtrace(AB)$? That is at least not obvious for me.
answered Apr 8 at 20:14
JonasJonas
415312
$endgroup$
$begingroup$
"Z is your matrix-valued Lagrange multiplier" Makes perfect sense. Thank you.
$endgroup$
– John Deterious
Apr 8 at 23:21
add a comment |
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1 Answer
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oldest
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oldest
votes
$begingroup$
I think, you need the trace in this setting, since the constraint is not scalar, but matrix-valued. The $Z$ is your matrix-valued Lagrange multiplier, which cannot be scalar in this setting.
I have written down a the Lagrange approach for such very general equality constraints:
beginalign
max_x in X f(x) \ texts.t. g(x) = 0,
endalign
where $f:X rightarrow mathbbR$, and $g: X rightarrow Y$, where $X$ and $Y$ are Hilbert spaces (like $mathbbR^k$, $mathcalL^2$,...).
The Lagrange function is given by
$$L(x, Lambda) = f(x) + langle Lambda, g(x) rangle_Y,$$
where $Lambda in Y$ is the Lagrange multiplier, and $langle cdot, cdot rangle_Y$ is the inner product of $Y$.
The question is: Does the set of matrices that is some superset of the image of $BC-I$ in your notation, form a Hilbert space $Y$ with the trace of the product of two matrices being their inner product; i.e. $langle A, B rangle_X := mathrmtrace(AB)$? That is at least not obvious for me.
answered Apr 8 at 20:14
JonasJonas
415312
$endgroup$
$begingroup$
"Z is your matrix-valued Lagrange multiplier" Makes perfect sense. Thank you.
$endgroup$
– John Deterious
Apr 8 at 23:21
add a comment |
$begingroup$
I think, you need the trace in this setting, since the constraint is not scalar, but matrix-valued. The $Z$ is your matrix-valued Lagrange multiplier, which cannot be scalar in this setting.
I have written down a the Lagrange approach for such very general equality constraints:
beginalign
max_x in X f(x) \ texts.t. g(x) = 0,
endalign
where $f:X rightarrow mathbbR$, and $g: X rightarrow Y$, where $X$ and $Y$ are Hilbert spaces (like $mathbbR^k$, $mathcalL^2$,...).
The Lagrange function is given by
$$L(x, Lambda) = f(x) + langle Lambda, g(x) rangle_Y,$$
where $Lambda in Y$ is the Lagrange multiplier, and $langle cdot, cdot rangle_Y$ is the inner product of $Y$.
The question is: Does the set of matrices that is some superset of the image of $BC-I$ in your notation, form a Hilbert space $Y$ with the trace of the product of two matrices being their inner product; i.e. $langle A, B rangle_X := mathrmtrace(AB)$? That is at least not obvious for me.
answered Apr 8 at 20:14
JonasJonas
415312
$endgroup$
$begingroup$
"Z is your matrix-valued Lagrange multiplier" Makes perfect sense. Thank you.
$endgroup$
– John Deterious
Apr 8 at 23:21
add a comment |
$begingroup$
I think, you need the trace in this setting, since the constraint is not scalar, but matrix-valued. The $Z$ is your matrix-valued Lagrange multiplier, which cannot be scalar in this setting.
I have written down a the Lagrange approach for such very general equality constraints:
beginalign
max_x in X f(x) \ texts.t. g(x) = 0,
endalign
where $f:X rightarrow mathbbR$, and $g: X rightarrow Y$, where $X$ and $Y$ are Hilbert spaces (like $mathbbR^k$, $mathcalL^2$,...).
The Lagrange function is given by
$$L(x, Lambda) = f(x) + langle Lambda, g(x) rangle_Y,$$
where $Lambda in Y$ is the Lagrange multiplier, and $langle cdot, cdot rangle_Y$ is the inner product of $Y$.
The question is: Does the set of matrices that is some superset of the image of $BC-I$ in your notation, form a Hilbert space $Y$ with the trace of the product of two matrices being their inner product; i.e. $langle A, B rangle_X := mathrmtrace(AB)$? That is at least not obvious for me.
answered Apr 8 at 20:14
JonasJonas
415312
$endgroup$
I think, you need the trace in this setting, since the constraint is not scalar, but matrix-valued. The $Z$ is your matrix-valued Lagrange multiplier, which cannot be scalar in this setting.
I have written down a the Lagrange approach for such very general equality constraints:
beginalign
max_x in X f(x) \ texts.t. g(x) = 0,
endalign
where $f:X rightarrow mathbbR$, and $g: X rightarrow Y$, where $X$ and $Y$ are Hilbert spaces (like $mathbbR^k$, $mathcalL^2$,...).
The Lagrange function is given by
$$L(x, Lambda) = f(x) + langle Lambda, g(x) rangle_Y,$$
where $Lambda in Y$ is the Lagrange multiplier, and $langle cdot, cdot rangle_Y$ is the inner product of $Y$.
The question is: Does the set of matrices that is some superset of the image of $BC-I$ in your notation, form a Hilbert space $Y$ with the trace of the product of two matrices being their inner product; i.e. $langle A, B rangle_X := mathrmtrace(AB)$? That is at least not obvious for me.
answered Apr 8 at 20:14
JonasJonas
415312
answered Apr 8 at 20:14
JonasJonas
415312
answered Apr 8 at 20:14
JonasJonas
415312
answered Apr 8 at 20:14
JonasJonas
415312
415312
$begingroup$
"Z is your matrix-valued Lagrange multiplier" Makes perfect sense. Thank you.
$endgroup$
– John Deterious
Apr 8 at 23:21
add a comment |
$begingroup$
"Z is your matrix-valued Lagrange multiplier" Makes perfect sense. Thank you.
$endgroup$
– John Deterious
Apr 8 at 23:21
$begingroup$
"Z is your matrix-valued Lagrange multiplier" Makes perfect sense. Thank you.
$endgroup$
– John Deterious
Apr 8 at 23:21
$begingroup$
"Z is your matrix-valued Lagrange multiplier" Makes perfect sense. Thank you.
$endgroup$
– John Deterious
Apr 8 at 23:21
add a comment |
John Deterious is a new contributor. Be nice, and check out our Code of Conduct.
John Deterious is a new contributor. Be nice, and check out our Code of Conduct.
John Deterious is a new contributor. Be nice, and check out our Code of Conduct.
John Deterious is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
PS: that wasn't the exact problem, just a mock, but it conveys the idea.
$endgroup$
– John Deterious
Apr 8 at 19:24
$begingroup$
Yeah, the $Z$ is probably a $2$.
$endgroup$
– Mark Fischler
Apr 8 at 19:24
$begingroup$
@MarkFischler No, it is a symmetric matrix. Somehow it helps.
$endgroup$
– John Deterious
Apr 8 at 19:25
$begingroup$
And the condition that $BC = I$ can't be what you meant, because this becomes finding $min mbox tr(AD)$.
$endgroup$
– Mark Fischler
Apr 8 at 19:26
$begingroup$
@MarkFischler This is just a toy example, you are correct. But, still, regardless of that point, what's the answer?
$endgroup$
– John Deterious
Apr 8 at 19:27