How to show polyhedral cone of nonnegative vectors contains finitely generated cone?
$begingroup$
Let $P={x in mathbb{R}^n mid Ax geq b, x geq 0 }$ be a nonempty polyhedron for matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^m$.
According to Minkowski-Weyl theorem $P$ can be written as
$$
P=text{conv}(v_1,cdots,v_p)+ text{cone}(d_1,cdots,d_l)
$$
for some $v_i in mathbb{R}^n$ and $d_j in mathbb{R}^n$.
Let $C={x in mathbb{R}^n mid Ax geq 0, x geq 0 }$.
Show that $text{cone}(d_1,cdots,d_l) subseteq C$.
The thing that that I cannot cope with is how to connect the finite number $l$ that can be any natural number with dimension of the matrix $A$.
I tried the following:
Let $z in text{cone}(d_1,cdots,d_l)$, so there exist non-negative $mu_i$'s such that
$$
z= sum_{i=1}^l mu_id_i
$$
where $mu_1,mu_2,cdots,mu_l geq 0$.
We can write $z$ as the following:
$$
z=
begin{bmatrix}
d_1 & d_2 & cdots & d_l
end{bmatrix}
begin{bmatrix}
mu_1 \
mu_2 \
cdots \
mu_l
end{bmatrix}
$$
Now, we should come up with an $m times n$ matrix $A$ for which we have $Az geq 0$ and $z geq 0$ to prove the claim. But the problem is we do not have $z geq 0$ necessarily.
convex-analysis convex-geometry
asked Dec 14 '18 at 6:43
SepideSepide
3308
$endgroup$
add a comment |
$begingroup$
Let $P={x in mathbb{R}^n mid Ax geq b, x geq 0 }$ be a nonempty polyhedron for matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^m$.
According to Minkowski-Weyl theorem $P$ can be written as
$$
P=text{conv}(v_1,cdots,v_p)+ text{cone}(d_1,cdots,d_l)
$$
for some $v_i in mathbb{R}^n$ and $d_j in mathbb{R}^n$.
Let $C={x in mathbb{R}^n mid Ax geq 0, x geq 0 }$.
Show that $text{cone}(d_1,cdots,d_l) subseteq C$.
The thing that that I cannot cope with is how to connect the finite number $l$ that can be any natural number with dimension of the matrix $A$.
I tried the following:
Let $z in text{cone}(d_1,cdots,d_l)$, so there exist non-negative $mu_i$'s such that
$$
z= sum_{i=1}^l mu_id_i
$$
where $mu_1,mu_2,cdots,mu_l geq 0$.
We can write $z$ as the following:
$$
z=
begin{bmatrix}
d_1 & d_2 & cdots & d_l
end{bmatrix}
begin{bmatrix}
mu_1 \
mu_2 \
cdots \
mu_l
end{bmatrix}
$$
Now, we should come up with an $m times n$ matrix $A$ for which we have $Az geq 0$ and $z geq 0$ to prove the claim. But the problem is we do not have $z geq 0$ necessarily.
convex-analysis convex-geometry
asked Dec 14 '18 at 6:43
SepideSepide
3308
$endgroup$
$begingroup$
The statement is not true for any choice of $v_i$, so you need to chose them appropriately.
$endgroup$
– LinAlg
Dec 14 '18 at 22:27
$begingroup$
@ LinAlg: The statement is true because it says for some $v_i$ and $d_j$. Also, it does not say what $p$ and $l$ are. Maybe I am wrong who am using wrong number for $l$.
$endgroup$
– Sepide
Dec 15 '18 at 0:22
$begingroup$
a proof by contradiction is easier here: what if $z$ is not in $C$?
$endgroup$
– LinAlg
Dec 15 '18 at 0:59
$begingroup$
@LinAlg: How we can do that?
$endgroup$
– Sepide
Dec 15 '18 at 4:13
$begingroup$
if $z$ does not satisfy $Azgeq 0$ or $zgeq 0$, moving in the direction of $z$ will violate a constraint of $P$
$endgroup$
– LinAlg
Dec 15 '18 at 16:19
add a comment |
$begingroup$
Let $P={x in mathbb{R}^n mid Ax geq b, x geq 0 }$ be a nonempty polyhedron for matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^m$.
According to Minkowski-Weyl theorem $P$ can be written as
$$
P=text{conv}(v_1,cdots,v_p)+ text{cone}(d_1,cdots,d_l)
$$
for some $v_i in mathbb{R}^n$ and $d_j in mathbb{R}^n$.
Let $C={x in mathbb{R}^n mid Ax geq 0, x geq 0 }$.
Show that $text{cone}(d_1,cdots,d_l) subseteq C$.
The thing that that I cannot cope with is how to connect the finite number $l$ that can be any natural number with dimension of the matrix $A$.
I tried the following:
Let $z in text{cone}(d_1,cdots,d_l)$, so there exist non-negative $mu_i$'s such that
$$
z= sum_{i=1}^l mu_id_i
$$
where $mu_1,mu_2,cdots,mu_l geq 0$.
We can write $z$ as the following:
$$
z=
begin{bmatrix}
d_1 & d_2 & cdots & d_l
end{bmatrix}
begin{bmatrix}
mu_1 \
mu_2 \
cdots \
mu_l
end{bmatrix}
$$
Now, we should come up with an $m times n$ matrix $A$ for which we have $Az geq 0$ and $z geq 0$ to prove the claim. But the problem is we do not have $z geq 0$ necessarily.
convex-analysis convex-geometry
asked Dec 14 '18 at 6:43
SepideSepide
3308
$endgroup$
Let $P={x in mathbb{R}^n mid Ax geq b, x geq 0 }$ be a nonempty polyhedron for matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^m$.
According to Minkowski-Weyl theorem $P$ can be written as
$$
P=text{conv}(v_1,cdots,v_p)+ text{cone}(d_1,cdots,d_l)
$$
for some $v_i in mathbb{R}^n$ and $d_j in mathbb{R}^n$.
Let $C={x in mathbb{R}^n mid Ax geq 0, x geq 0 }$.
Show that $text{cone}(d_1,cdots,d_l) subseteq C$.
The thing that that I cannot cope with is how to connect the finite number $l$ that can be any natural number with dimension of the matrix $A$.
I tried the following:
Let $z in text{cone}(d_1,cdots,d_l)$, so there exist non-negative $mu_i$'s such that
$$
z= sum_{i=1}^l mu_id_i
$$
where $mu_1,mu_2,cdots,mu_l geq 0$.
We can write $z$ as the following:
$$
z=
begin{bmatrix}
d_1 & d_2 & cdots & d_l
end{bmatrix}
begin{bmatrix}
mu_1 \
mu_2 \
cdots \
mu_l
end{bmatrix}
$$
Now, we should come up with an $m times n$ matrix $A$ for which we have $Az geq 0$ and $z geq 0$ to prove the claim. But the problem is we do not have $z geq 0$ necessarily.
convex-analysis convex-geometry
convex-analysis convex-geometry
asked Dec 14 '18 at 6:43
SepideSepide
3308
asked Dec 14 '18 at 6:43
SepideSepide
3308
edited Dec 15 '18 at 0:23
Sepide
asked Dec 14 '18 at 6:43
SepideSepide
3308
asked Dec 14 '18 at 6:43
SepideSepide
3308
asked Dec 14 '18 at 6:43
SepideSepide
3308
3308
$begingroup$
The statement is not true for any choice of $v_i$, so you need to chose them appropriately.
$endgroup$
– LinAlg
Dec 14 '18 at 22:27
$begingroup$
@ LinAlg: The statement is true because it says for some $v_i$ and $d_j$. Also, it does not say what $p$ and $l$ are. Maybe I am wrong who am using wrong number for $l$.
$endgroup$
– Sepide
Dec 15 '18 at 0:22
$begingroup$
a proof by contradiction is easier here: what if $z$ is not in $C$?
$endgroup$
– LinAlg
Dec 15 '18 at 0:59
$begingroup$
@LinAlg: How we can do that?
$endgroup$
– Sepide
Dec 15 '18 at 4:13
$begingroup$
if $z$ does not satisfy $Azgeq 0$ or $zgeq 0$, moving in the direction of $z$ will violate a constraint of $P$
$endgroup$
– LinAlg
Dec 15 '18 at 16:19
add a comment |
$begingroup$
The statement is not true for any choice of $v_i$, so you need to chose them appropriately.
$endgroup$
– LinAlg
Dec 14 '18 at 22:27
$begingroup$
@ LinAlg: The statement is true because it says for some $v_i$ and $d_j$. Also, it does not say what $p$ and $l$ are. Maybe I am wrong who am using wrong number for $l$.
$endgroup$
– Sepide
Dec 15 '18 at 0:22
$begingroup$
a proof by contradiction is easier here: what if $z$ is not in $C$?
$endgroup$
– LinAlg
Dec 15 '18 at 0:59
$begingroup$
@LinAlg: How we can do that?
$endgroup$
– Sepide
Dec 15 '18 at 4:13
$begingroup$
if $z$ does not satisfy $Azgeq 0$ or $zgeq 0$, moving in the direction of $z$ will violate a constraint of $P$
$endgroup$
– LinAlg
Dec 15 '18 at 16:19
$begingroup$
The statement is not true for any choice of $v_i$, so you need to chose them appropriately.
$endgroup$
– LinAlg
Dec 14 '18 at 22:27
$begingroup$
The statement is not true for any choice of $v_i$, so you need to chose them appropriately.
$endgroup$
– LinAlg
Dec 14 '18 at 22:27
$begingroup$
@ LinAlg: The statement is true because it says for some $v_i$ and $d_j$. Also, it does not say what $p$ and $l$ are. Maybe I am wrong who am using wrong number for $l$.
$endgroup$
– Sepide
Dec 15 '18 at 0:22
$begingroup$
@ LinAlg: The statement is true because it says for some $v_i$ and $d_j$. Also, it does not say what $p$ and $l$ are. Maybe I am wrong who am using wrong number for $l$.
$endgroup$
– Sepide
Dec 15 '18 at 0:22
$begingroup$
a proof by contradiction is easier here: what if $z$ is not in $C$?
$endgroup$
– LinAlg
Dec 15 '18 at 0:59
$begingroup$
a proof by contradiction is easier here: what if $z$ is not in $C$?
$endgroup$
– LinAlg
Dec 15 '18 at 0:59
$begingroup$
@LinAlg: How we can do that?
$endgroup$
– Sepide
Dec 15 '18 at 4:13
$begingroup$
@LinAlg: How we can do that?
$endgroup$
– Sepide
Dec 15 '18 at 4:13
$begingroup$
if $z$ does not satisfy $Azgeq 0$ or $zgeq 0$, moving in the direction of $z$ will violate a constraint of $P$
$endgroup$
– LinAlg
Dec 15 '18 at 16:19
$begingroup$
if $z$ does not satisfy $Azgeq 0$ or $zgeq 0$, moving in the direction of $z$ will violate a constraint of $P$
$endgroup$
– LinAlg
Dec 15 '18 at 16:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We know that $P$ can be written as
$$
P=operatorname{conv}(v_1,cdots,v_p)+ operatorname{cone}(d_1,cdots,d_l)=V+D.
$$
The set $D$ is a cone, hence, for every $vin V$ and $din D$ we have that $v+tdin P$, $forall tge 0$. That is
$$
A(v+td)ge b,quad v+tdge 0,quadforall tge 0.
$$
Now divide by $t$ and let $tto +infty$
begin{align}
frac{1}{t}Av+Adgefrac{1}{t}bquad&Rightarrowquad Adge 0,\
frac{1}{t}v+dge 0 quad&Rightarrowquad dge 0.
end{align}
Therefore, every $din D$ belongs to $C$.
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
$endgroup$
$begingroup$
Could you help me show the reverse? I mean $ C subseteq text{cone}(d_1,cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x geq 0$?
$endgroup$
– Sepide
Dec 15 '18 at 18:32
$begingroup$
I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/…
$endgroup$
– Sepide
Dec 15 '18 at 19:54
add a comment |
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$begingroup$
We know that $P$ can be written as
$$
P=operatorname{conv}(v_1,cdots,v_p)+ operatorname{cone}(d_1,cdots,d_l)=V+D.
$$
The set $D$ is a cone, hence, for every $vin V$ and $din D$ we have that $v+tdin P$, $forall tge 0$. That is
$$
A(v+td)ge b,quad v+tdge 0,quadforall tge 0.
$$
Now divide by $t$ and let $tto +infty$
begin{align}
frac{1}{t}Av+Adgefrac{1}{t}bquad&Rightarrowquad Adge 0,\
frac{1}{t}v+dge 0 quad&Rightarrowquad dge 0.
end{align}
Therefore, every $din D$ belongs to $C$.
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
$endgroup$
$begingroup$
Could you help me show the reverse? I mean $ C subseteq text{cone}(d_1,cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x geq 0$?
$endgroup$
– Sepide
Dec 15 '18 at 18:32
$begingroup$
I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/…
$endgroup$
– Sepide
Dec 15 '18 at 19:54
add a comment |
$begingroup$
We know that $P$ can be written as
$$
P=operatorname{conv}(v_1,cdots,v_p)+ operatorname{cone}(d_1,cdots,d_l)=V+D.
$$
The set $D$ is a cone, hence, for every $vin V$ and $din D$ we have that $v+tdin P$, $forall tge 0$. That is
$$
A(v+td)ge b,quad v+tdge 0,quadforall tge 0.
$$
Now divide by $t$ and let $tto +infty$
begin{align}
frac{1}{t}Av+Adgefrac{1}{t}bquad&Rightarrowquad Adge 0,\
frac{1}{t}v+dge 0 quad&Rightarrowquad dge 0.
end{align}
Therefore, every $din D$ belongs to $C$.
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
$endgroup$
$begingroup$
Could you help me show the reverse? I mean $ C subseteq text{cone}(d_1,cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x geq 0$?
$endgroup$
– Sepide
Dec 15 '18 at 18:32
$begingroup$
I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/…
$endgroup$
– Sepide
Dec 15 '18 at 19:54
add a comment |
$begingroup$
We know that $P$ can be written as
$$
P=operatorname{conv}(v_1,cdots,v_p)+ operatorname{cone}(d_1,cdots,d_l)=V+D.
$$
The set $D$ is a cone, hence, for every $vin V$ and $din D$ we have that $v+tdin P$, $forall tge 0$. That is
$$
A(v+td)ge b,quad v+tdge 0,quadforall tge 0.
$$
Now divide by $t$ and let $tto +infty$
begin{align}
frac{1}{t}Av+Adgefrac{1}{t}bquad&Rightarrowquad Adge 0,\
frac{1}{t}v+dge 0 quad&Rightarrowquad dge 0.
end{align}
Therefore, every $din D$ belongs to $C$.
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
$endgroup$
We know that $P$ can be written as
$$
P=operatorname{conv}(v_1,cdots,v_p)+ operatorname{cone}(d_1,cdots,d_l)=V+D.
$$
The set $D$ is a cone, hence, for every $vin V$ and $din D$ we have that $v+tdin P$, $forall tge 0$. That is
$$
A(v+td)ge b,quad v+tdge 0,quadforall tge 0.
$$
Now divide by $t$ and let $tto +infty$
begin{align}
frac{1}{t}Av+Adgefrac{1}{t}bquad&Rightarrowquad Adge 0,\
frac{1}{t}v+dge 0 quad&Rightarrowquad dge 0.
end{align}
Therefore, every $din D$ belongs to $C$.
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
answered Dec 15 '18 at 12:01
A.Γ.A.Γ.
22.7k32656
22.7k32656
$begingroup$
Could you help me show the reverse? I mean $ C subseteq text{cone}(d_1,cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x geq 0$?
$endgroup$
– Sepide
Dec 15 '18 at 18:32
$begingroup$
I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/…
$endgroup$
– Sepide
Dec 15 '18 at 19:54
add a comment |
$begingroup$
Could you help me show the reverse? I mean $ C subseteq text{cone}(d_1,cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x geq 0$?
$endgroup$
– Sepide
Dec 15 '18 at 18:32
$begingroup$
I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/…
$endgroup$
– Sepide
Dec 15 '18 at 19:54
$begingroup$
Could you help me show the reverse? I mean $ C subseteq text{cone}(d_1,cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x geq 0$?
$endgroup$
– Sepide
Dec 15 '18 at 18:32
$begingroup$
Could you help me show the reverse? I mean $ C subseteq text{cone}(d_1,cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x geq 0$?
$endgroup$
– Sepide
Dec 15 '18 at 18:32
$begingroup$
I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/…
$endgroup$
– Sepide
Dec 15 '18 at 19:54
$begingroup$
I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/…
$endgroup$
– Sepide
Dec 15 '18 at 19:54
add a comment |
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$begingroup$
The statement is not true for any choice of $v_i$, so you need to chose them appropriately.
$endgroup$
– LinAlg
Dec 14 '18 at 22:27
$begingroup$
@ LinAlg: The statement is true because it says for some $v_i$ and $d_j$. Also, it does not say what $p$ and $l$ are. Maybe I am wrong who am using wrong number for $l$.
$endgroup$
– Sepide
Dec 15 '18 at 0:22
$begingroup$
a proof by contradiction is easier here: what if $z$ is not in $C$?
$endgroup$
– LinAlg
Dec 15 '18 at 0:59
$begingroup$
@LinAlg: How we can do that?
$endgroup$
– Sepide
Dec 15 '18 at 4:13
$begingroup$
if $z$ does not satisfy $Azgeq 0$ or $zgeq 0$, moving in the direction of $z$ will violate a constraint of $P$
$endgroup$
– LinAlg
Dec 15 '18 at 16:19