Prove that the normal cone $N_{text{gph}(f)}$ of the graph of the affine function $f$ has the given form.
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GIVEN
Let $f : mathbb{R}^n longrightarrow mathbb{R}^m$ be the affine function defined by $f(x) = Mx + b$ where $M$ is an $m times n$ matrix and $b$ is a vector in $mathbb{R}^m$. Prove that for $(x_0,y_0) in text{gph}(f)$:
$$
N_{text{gph}(f)}(x_0,y_0)=big{ (u,v)inmathbb{R}^n times mathbb{R}^m:u+M^Tv=0 big}.
$$
USEFUL DEFINITION
Normal Cone
The normal cone of a nonempty, closed, and convex set $K$ at a point $x_0 in K$ is:
$$
N_K(x_0)=big{ z:langle z, ;x-x_0 rangle leq 0,; forall x in Kbig}
$$
ATTEMPT
By applying the above definitions I was able to reach
$$
langle u + M^Tv , ; x - x_0rangle leq 0
$$
I am not sure how to prove that $u + M^Tv = 0$.
One thing that might help: the subdifferential of $f$ is $partial f = { nabla f } = { M^T }$.
Any help is greatly appreciated.
linear-algebra vector-spaces convex-analysis convex-optimization convex-geometry
asked Nov 23 at 11:43
ex.nihil
17910
add a comment |
up vote
1
down vote
favorite
GIVEN
Let $f : mathbb{R}^n longrightarrow mathbb{R}^m$ be the affine function defined by $f(x) = Mx + b$ where $M$ is an $m times n$ matrix and $b$ is a vector in $mathbb{R}^m$. Prove that for $(x_0,y_0) in text{gph}(f)$:
$$
N_{text{gph}(f)}(x_0,y_0)=big{ (u,v)inmathbb{R}^n times mathbb{R}^m:u+M^Tv=0 big}.
$$
USEFUL DEFINITION
Normal Cone
The normal cone of a nonempty, closed, and convex set $K$ at a point $x_0 in K$ is:
$$
N_K(x_0)=big{ z:langle z, ;x-x_0 rangle leq 0,; forall x in Kbig}
$$
ATTEMPT
By applying the above definitions I was able to reach
$$
langle u + M^Tv , ; x - x_0rangle leq 0
$$
I am not sure how to prove that $u + M^Tv = 0$.
One thing that might help: the subdifferential of $f$ is $partial f = { nabla f } = { M^T }$.
Any help is greatly appreciated.
linear-algebra vector-spaces convex-analysis convex-optimization convex-geometry
asked Nov 23 at 11:43
ex.nihil
17910
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
GIVEN
Let $f : mathbb{R}^n longrightarrow mathbb{R}^m$ be the affine function defined by $f(x) = Mx + b$ where $M$ is an $m times n$ matrix and $b$ is a vector in $mathbb{R}^m$. Prove that for $(x_0,y_0) in text{gph}(f)$:
$$
N_{text{gph}(f)}(x_0,y_0)=big{ (u,v)inmathbb{R}^n times mathbb{R}^m:u+M^Tv=0 big}.
$$
USEFUL DEFINITION
Normal Cone
The normal cone of a nonempty, closed, and convex set $K$ at a point $x_0 in K$ is:
$$
N_K(x_0)=big{ z:langle z, ;x-x_0 rangle leq 0,; forall x in Kbig}
$$
ATTEMPT
By applying the above definitions I was able to reach
$$
langle u + M^Tv , ; x - x_0rangle leq 0
$$
I am not sure how to prove that $u + M^Tv = 0$.
One thing that might help: the subdifferential of $f$ is $partial f = { nabla f } = { M^T }$.
Any help is greatly appreciated.
linear-algebra vector-spaces convex-analysis convex-optimization convex-geometry
asked Nov 23 at 11:43
ex.nihil
17910
GIVEN
Let $f : mathbb{R}^n longrightarrow mathbb{R}^m$ be the affine function defined by $f(x) = Mx + b$ where $M$ is an $m times n$ matrix and $b$ is a vector in $mathbb{R}^m$. Prove that for $(x_0,y_0) in text{gph}(f)$:
$$
N_{text{gph}(f)}(x_0,y_0)=big{ (u,v)inmathbb{R}^n times mathbb{R}^m:u+M^Tv=0 big}.
$$
USEFUL DEFINITION
Normal Cone
The normal cone of a nonempty, closed, and convex set $K$ at a point $x_0 in K$ is:
$$
N_K(x_0)=big{ z:langle z, ;x-x_0 rangle leq 0,; forall x in Kbig}
$$
ATTEMPT
By applying the above definitions I was able to reach
$$
langle u + M^Tv , ; x - x_0rangle leq 0
$$
I am not sure how to prove that $u + M^Tv = 0$.
One thing that might help: the subdifferential of $f$ is $partial f = { nabla f } = { M^T }$.
Any help is greatly appreciated.
linear-algebra vector-spaces convex-analysis convex-optimization convex-geometry
linear-algebra vector-spaces convex-analysis convex-optimization convex-geometry
asked Nov 23 at 11:43
ex.nihil
17910
asked Nov 23 at 11:43
ex.nihil
17910
asked Nov 23 at 11:43
ex.nihil
17910
asked Nov 23 at 11:43
ex.nihil
17910
asked Nov 23 at 11:43
ex.nihil
17910
17910
add a comment |
add a comment |
1 Answer
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1
down vote
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You are on a good way towards proving $u+M^top v=0$.
If you have
$$
langle u+M^top v,;x-x_0rangle leq 0
$$
for all $xinmathbb R^n$, then (by substituting $x$ with $x+x_0$) it follows that
$$
langle u+M^top v,;xrangle leq 0
$$
for all $xinmathbb R^n$.
By substituting $x$ with $-x$, we get
$$
langle u+M^top v,;xrangle =0
$$
for all $xinmathbb R^n$.
This implies (by linear algebra) that $u+M^top v=0$.
This would complete the "$subset$" part of the proof.
The other "$supset$" part is easier in my opinion, so you should give it a try.
answered Nov 23 at 12:06
supinf
5,9501027
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
up vote
1
down vote
accepted
You are on a good way towards proving $u+M^top v=0$.
If you have
$$
langle u+M^top v,;x-x_0rangle leq 0
$$
for all $xinmathbb R^n$, then (by substituting $x$ with $x+x_0$) it follows that
$$
langle u+M^top v,;xrangle leq 0
$$
for all $xinmathbb R^n$.
By substituting $x$ with $-x$, we get
$$
langle u+M^top v,;xrangle =0
$$
for all $xinmathbb R^n$.
This implies (by linear algebra) that $u+M^top v=0$.
This would complete the "$subset$" part of the proof.
The other "$supset$" part is easier in my opinion, so you should give it a try.
answered Nov 23 at 12:06
supinf
5,9501027
add a comment |
up vote
1
down vote
accepted
You are on a good way towards proving $u+M^top v=0$.
If you have
$$
langle u+M^top v,;x-x_0rangle leq 0
$$
for all $xinmathbb R^n$, then (by substituting $x$ with $x+x_0$) it follows that
$$
langle u+M^top v,;xrangle leq 0
$$
for all $xinmathbb R^n$.
By substituting $x$ with $-x$, we get
$$
langle u+M^top v,;xrangle =0
$$
for all $xinmathbb R^n$.
This implies (by linear algebra) that $u+M^top v=0$.
This would complete the "$subset$" part of the proof.
The other "$supset$" part is easier in my opinion, so you should give it a try.
answered Nov 23 at 12:06
supinf
5,9501027
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are on a good way towards proving $u+M^top v=0$.
If you have
$$
langle u+M^top v,;x-x_0rangle leq 0
$$
for all $xinmathbb R^n$, then (by substituting $x$ with $x+x_0$) it follows that
$$
langle u+M^top v,;xrangle leq 0
$$
for all $xinmathbb R^n$.
By substituting $x$ with $-x$, we get
$$
langle u+M^top v,;xrangle =0
$$
for all $xinmathbb R^n$.
This implies (by linear algebra) that $u+M^top v=0$.
This would complete the "$subset$" part of the proof.
The other "$supset$" part is easier in my opinion, so you should give it a try.
answered Nov 23 at 12:06
supinf
5,9501027
You are on a good way towards proving $u+M^top v=0$.
If you have
$$
langle u+M^top v,;x-x_0rangle leq 0
$$
for all $xinmathbb R^n$, then (by substituting $x$ with $x+x_0$) it follows that
$$
langle u+M^top v,;xrangle leq 0
$$
for all $xinmathbb R^n$.
By substituting $x$ with $-x$, we get
$$
langle u+M^top v,;xrangle =0
$$
for all $xinmathbb R^n$.
This implies (by linear algebra) that $u+M^top v=0$.
This would complete the "$subset$" part of the proof.
The other "$supset$" part is easier in my opinion, so you should give it a try.
answered Nov 23 at 12:06
supinf
5,9501027
answered Nov 23 at 12:06
supinf
5,9501027
answered Nov 23 at 12:06
supinf
5,9501027
answered Nov 23 at 12:06
supinf
5,9501027
5,9501027
add a comment |
add a comment |
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