Mixing
Find at least one solution to matrix inequality
I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$
for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.
Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!
linear-programming matrix-equations geometric-inequalities
asked Nov 21 '18 at 17:16
MajinSaha
405
add a comment |
I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$
for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.
Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!
linear-programming matrix-equations geometric-inequalities
asked Nov 21 '18 at 17:16
MajinSaha
405
You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 '18 at 17:40
@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 '18 at 17:41
add a comment |
I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$
for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.
Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!
linear-programming matrix-equations geometric-inequalities
asked Nov 21 '18 at 17:16
MajinSaha
405
I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$
for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.
Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!
linear-programming matrix-equations geometric-inequalities
linear-programming matrix-equations geometric-inequalities
asked Nov 21 '18 at 17:16
MajinSaha
405
asked Nov 21 '18 at 17:16
MajinSaha
405
asked Nov 21 '18 at 17:16
MajinSaha
405
asked Nov 21 '18 at 17:16
MajinSaha
405
asked Nov 21 '18 at 17:16
MajinSaha
405
405
You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 '18 at 17:40
@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 '18 at 17:41
add a comment |
You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 '18 at 17:40
@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 '18 at 17:41
You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 '18 at 17:40
You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 '18 at 17:40
@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 '18 at 17:41
@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 '18 at 17:41
add a comment |
2 Answers
2
active
oldest
votes
Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:
An infeasible interior point method. This solves
$$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$A feasible interior point method. Consider the extended linear optimization problem:
$$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.With the two phase simplex method in which you only need the first phase. This is essentially solving
$$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 '18 at 23:28
@MajinSaha did it answer your question?
– LinAlg
Dec 7 '18 at 17:43
Yes. I wish there was a shorter way, but I guess that's the best one out there.
– MajinSaha
Dec 9 '18 at 4:12
add a comment |
Linear programming is your method of choice.
A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.
Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
answered Nov 21 '18 at 17:44
Andreas
7,8281037
Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 '18 at 19:01
This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 '18 at 21:56
I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 '18 at 23:26
Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
Nov 22 '18 at 9:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:
An infeasible interior point method. This solves
$$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$A feasible interior point method. Consider the extended linear optimization problem:
$$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.With the two phase simplex method in which you only need the first phase. This is essentially solving
$$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 '18 at 23:28
@MajinSaha did it answer your question?
– LinAlg
Dec 7 '18 at 17:43
Yes. I wish there was a shorter way, but I guess that's the best one out there.
– MajinSaha
Dec 9 '18 at 4:12
add a comment |
Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:
An infeasible interior point method. This solves
$$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$A feasible interior point method. Consider the extended linear optimization problem:
$$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.With the two phase simplex method in which you only need the first phase. This is essentially solving
$$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 '18 at 23:28
@MajinSaha did it answer your question?
– LinAlg
Dec 7 '18 at 17:43
Yes. I wish there was a shorter way, but I guess that's the best one out there.
– MajinSaha
Dec 9 '18 at 4:12
add a comment |
Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:
An infeasible interior point method. This solves
$$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$A feasible interior point method. Consider the extended linear optimization problem:
$$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.With the two phase simplex method in which you only need the first phase. This is essentially solving
$$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:
An infeasible interior point method. This solves
$$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$A feasible interior point method. Consider the extended linear optimization problem:
$$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.With the two phase simplex method in which you only need the first phase. This is essentially solving
$$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
answered Nov 21 '18 at 21:54
LinAlg
8,7511521
8,7511521
Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 '18 at 23:28
@MajinSaha did it answer your question?
– LinAlg
Dec 7 '18 at 17:43
Yes. I wish there was a shorter way, but I guess that's the best one out there.
– MajinSaha
Dec 9 '18 at 4:12
add a comment |
Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 '18 at 23:28
@MajinSaha did it answer your question?
– LinAlg
Dec 7 '18 at 17:43
Yes. I wish there was a shorter way, but I guess that's the best one out there.
– MajinSaha
Dec 9 '18 at 4:12
Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 '18 at 23:28
Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 '18 at 23:28
@MajinSaha did it answer your question?
– LinAlg
Dec 7 '18 at 17:43
@MajinSaha did it answer your question?
– LinAlg
Dec 7 '18 at 17:43
Yes. I wish there was a shorter way, but I guess that's the best one out there.
– MajinSaha
Dec 9 '18 at 4:12
Yes. I wish there was a shorter way, but I guess that's the best one out there.
– MajinSaha
Dec 9 '18 at 4:12
add a comment |
Linear programming is your method of choice.
A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.
Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
answered Nov 21 '18 at 17:44
Andreas
7,8281037
Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 '18 at 19:01
This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 '18 at 21:56
I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 '18 at 23:26
Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
Nov 22 '18 at 9:33
add a comment |
Linear programming is your method of choice.
A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.
Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
answered Nov 21 '18 at 17:44
Andreas
7,8281037
Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 '18 at 19:01
This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 '18 at 21:56
I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 '18 at 23:26
Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
Nov 22 '18 at 9:33
add a comment |
Linear programming is your method of choice.
A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.
Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
answered Nov 21 '18 at 17:44
Andreas
7,8281037
Linear programming is your method of choice.
A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.
Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
answered Nov 21 '18 at 17:44
Andreas
7,8281037
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
edited Nov 21 '18 at 21:58
LinAlg
8,7511521
8,7511521
answered Nov 21 '18 at 17:44
Andreas
7,8281037
answered Nov 21 '18 at 17:44
Andreas
7,8281037
answered Nov 21 '18 at 17:44
Andreas
7,8281037
7,8281037
Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 '18 at 19:01
This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 '18 at 21:56
I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 '18 at 23:26
Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
Nov 22 '18 at 9:33
add a comment |
Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 '18 at 19:01
This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 '18 at 21:56
I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 '18 at 23:26
Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
Nov 22 '18 at 9:33
Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 '18 at 19:01
Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 '18 at 19:01
This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 '18 at 21:56
This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 '18 at 21:56
I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 '18 at 23:26
I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 '18 at 23:26
Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
Nov 22 '18 at 9:33
Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
Nov 22 '18 at 9:33
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You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 '18 at 17:40
@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 '18 at 17:41