Show that intersection of a polyhedron and affine set is a polyhedron.
$begingroup$
Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.
To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is
$$
L= {v: cv=0}
$$
$$
u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
$$
where $d=Cu$.
To show the claim I need to assume $x$ is in $P$, so
$$
Ax geq b
$$
Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.
Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
How can I proceed?
linear-algebra convex-analysis
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
$endgroup$
add a comment |
$begingroup$
Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.
To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is
$$
L= {v: cv=0}
$$
$$
u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
$$
where $d=Cu$.
To show the claim I need to assume $x$ is in $P$, so
$$
Ax geq b
$$
Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.
Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
How can I proceed?
linear-algebra convex-analysis
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
$endgroup$
add a comment |
$begingroup$
Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.
To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is
$$
L= {v: cv=0}
$$
$$
u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
$$
where $d=Cu$.
To show the claim I need to assume $x$ is in $P$, so
$$
Ax geq b
$$
Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.
Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
How can I proceed?
linear-algebra convex-analysis
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
$endgroup$
Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.
To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is
$$
L= {v: cv=0}
$$
$$
u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
$$
where $d=Cu$.
To show the claim I need to assume $x$ is in $P$, so
$$
Ax geq b
$$
Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.
Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
How can I proceed?
linear-algebra convex-analysis
linear-algebra convex-analysis
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
edited Dec 8 '18 at 4:49
Saeed
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
asked Dec 8 '18 at 0:00
SaeedSaeed
928310
928310
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
$endgroup$
$begingroup$
I think I got what you mean, just change the last $leq$ to $geq$.
$endgroup$
– Saeed
Dec 8 '18 at 5:02
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
$endgroup$
$begingroup$
I think I got what you mean, just change the last $leq$ to $geq$.
$endgroup$
– Saeed
Dec 8 '18 at 5:02
add a comment |
$begingroup$
Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
$endgroup$
$begingroup$
I think I got what you mean, just change the last $leq$ to $geq$.
$endgroup$
– Saeed
Dec 8 '18 at 5:02
add a comment |
$begingroup$
Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
$endgroup$
Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
answered Dec 8 '18 at 0:24
Theo BenditTheo Bendit
17.2k12149
17.2k12149
$begingroup$
I think I got what you mean, just change the last $leq$ to $geq$.
$endgroup$
– Saeed
Dec 8 '18 at 5:02
add a comment |
$begingroup$
I think I got what you mean, just change the last $leq$ to $geq$.
$endgroup$
– Saeed
Dec 8 '18 at 5:02
$begingroup$
I think I got what you mean, just change the last $leq$ to $geq$.
$endgroup$
– Saeed
Dec 8 '18 at 5:02
$begingroup$
I think I got what you mean, just change the last $leq$ to $geq$.
$endgroup$
– Saeed
Dec 8 '18 at 5:02
add a comment |
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