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If a linear function is nonnegative on a subspace, then the function is zero
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding nonnegative solutions to an underdetermined linear systemIs calling a linear-equation a linear-function, misnomer or completely wrong?Geometric interpretation of First order conditionFirst order condition in constrained optimization: Alternative characterization via normal conesEquivalent characterizations of the dual norm on finite dimensional vector spacesWidth of an ellipsoidDoes entry-wise non-negativity imply positive semidefiniteness?Every affine set can be expressed as the solution set of a system of linear equationswhat are some direct definitions of convexity for multivariable functions?How to show two different definitions of $alpha$-strongly convex are equivalent?
$begingroup$
Edit: Consider this solved. I used the wrong definition of linear function. I originally used the polynomial definition, $f(x) = acdot x + b$, not the linear map definition $f(alpha x) = alpha f(x), f(x + y) = f(x) + f(y)$.
This claim comes from "Convex Optimization" by Boyd and Vandenberghe of recharacterizing a convex optimization problem that has only equality constraints (Section 4.2).
Claim: If a linear function is nonnegative on a subspace then it must be zero on the subspace.
Context: It seems as if this is a general statement, but I've had surprising difficulty trying to prove it in general. The specific problem used is:
Minimize $f_0(x)$
subject to $Ax = b$ where $x in mathbbR^n$ and $A$ is some $n times n$ matrix
This claim is applied toward the function $nabla f_0(x)^T v = nabla f_0(x) cdot v geq 0$ which is a nonnegative linear function over $v in Nullspace(A) = N(A)$
Work so far: Let $f(x) = a cdot x + b$ be some nonnegative linear function over some subspace $S$. $0 in S rightarrow b geq 0$. Suppose there exists $x_0 in S$ such that $a cdot x_0 + b > 0$. By definition of the subspace, $-x_0 in S$ such that $-a cdot x + b geq 0 rightarrow b geq a cdot x_0 rightarrow 2b geq a cdot x_0 + b > 0$.
By the last relation, if $b = 0$, then contradiction occurs so $b > 0$. At this point, I get the gut feeling I'm doing something wrong.
linear-algebra
$endgroup$
add a comment |
$begingroup$
Edit: Consider this solved. I used the wrong definition of linear function. I originally used the polynomial definition, $f(x) = acdot x + b$, not the linear map definition $f(alpha x) = alpha f(x), f(x + y) = f(x) + f(y)$.
This claim comes from "Convex Optimization" by Boyd and Vandenberghe of recharacterizing a convex optimization problem that has only equality constraints (Section 4.2).
Claim: If a linear function is nonnegative on a subspace then it must be zero on the subspace.
Context: It seems as if this is a general statement, but I've had surprising difficulty trying to prove it in general. The specific problem used is:
Minimize $f_0(x)$
subject to $Ax = b$ where $x in mathbbR^n$ and $A$ is some $n times n$ matrix
This claim is applied toward the function $nabla f_0(x)^T v = nabla f_0(x) cdot v geq 0$ which is a nonnegative linear function over $v in Nullspace(A) = N(A)$
Work so far: Let $f(x) = a cdot x + b$ be some nonnegative linear function over some subspace $S$. $0 in S rightarrow b geq 0$. Suppose there exists $x_0 in S$ such that $a cdot x_0 + b > 0$. By definition of the subspace, $-x_0 in S$ such that $-a cdot x + b geq 0 rightarrow b geq a cdot x_0 rightarrow 2b geq a cdot x_0 + b > 0$.
By the last relation, if $b = 0$, then contradiction occurs so $b > 0$. At this point, I get the gut feeling I'm doing something wrong.
linear-algebra
$endgroup$
3
$begingroup$
You’re using the wrong definition of linear function: in this context $f$ is linear provided that $f(alpha v)=alpha f(v)$ for each scalar $alpha$ and vector $v$. Thus, if $f(v)$ were positive for some $v$ in the subspace, $f(-v)=-f(v)$ would be negative.
$endgroup$
– Brian M. Scott
Jan 9 '17 at 23:26
$begingroup$
Kill me. I facepalmed so hard. Thank you for pointing that out. The claim works out without much work. I'm used to referring that version as a "linear map" over linear function.
$endgroup$
– user64878
Jan 9 '17 at 23:30
add a comment |
$begingroup$
Edit: Consider this solved. I used the wrong definition of linear function. I originally used the polynomial definition, $f(x) = acdot x + b$, not the linear map definition $f(alpha x) = alpha f(x), f(x + y) = f(x) + f(y)$.
This claim comes from "Convex Optimization" by Boyd and Vandenberghe of recharacterizing a convex optimization problem that has only equality constraints (Section 4.2).
Claim: If a linear function is nonnegative on a subspace then it must be zero on the subspace.
Context: It seems as if this is a general statement, but I've had surprising difficulty trying to prove it in general. The specific problem used is:
Minimize $f_0(x)$
subject to $Ax = b$ where $x in mathbbR^n$ and $A$ is some $n times n$ matrix
This claim is applied toward the function $nabla f_0(x)^T v = nabla f_0(x) cdot v geq 0$ which is a nonnegative linear function over $v in Nullspace(A) = N(A)$
Work so far: Let $f(x) = a cdot x + b$ be some nonnegative linear function over some subspace $S$. $0 in S rightarrow b geq 0$. Suppose there exists $x_0 in S$ such that $a cdot x_0 + b > 0$. By definition of the subspace, $-x_0 in S$ such that $-a cdot x + b geq 0 rightarrow b geq a cdot x_0 rightarrow 2b geq a cdot x_0 + b > 0$.
By the last relation, if $b = 0$, then contradiction occurs so $b > 0$. At this point, I get the gut feeling I'm doing something wrong.
linear-algebra
$endgroup$
Edit: Consider this solved. I used the wrong definition of linear function. I originally used the polynomial definition, $f(x) = acdot x + b$, not the linear map definition $f(alpha x) = alpha f(x), f(x + y) = f(x) + f(y)$.
This claim comes from "Convex Optimization" by Boyd and Vandenberghe of recharacterizing a convex optimization problem that has only equality constraints (Section 4.2).
Claim: If a linear function is nonnegative on a subspace then it must be zero on the subspace.
Context: It seems as if this is a general statement, but I've had surprising difficulty trying to prove it in general. The specific problem used is:
Minimize $f_0(x)$
subject to $Ax = b$ where $x in mathbbR^n$ and $A$ is some $n times n$ matrix
This claim is applied toward the function $nabla f_0(x)^T v = nabla f_0(x) cdot v geq 0$ which is a nonnegative linear function over $v in Nullspace(A) = N(A)$
Work so far: Let $f(x) = a cdot x + b$ be some nonnegative linear function over some subspace $S$. $0 in S rightarrow b geq 0$. Suppose there exists $x_0 in S$ such that $a cdot x_0 + b > 0$. By definition of the subspace, $-x_0 in S$ such that $-a cdot x + b geq 0 rightarrow b geq a cdot x_0 rightarrow 2b geq a cdot x_0 + b > 0$.
By the last relation, if $b = 0$, then contradiction occurs so $b > 0$. At this point, I get the gut feeling I'm doing something wrong.
linear-algebra
linear-algebra
edited Jan 9 '17 at 23:32
asked Jan 9 '17 at 23:24
user64878
3
$begingroup$
You’re using the wrong definition of linear function: in this context $f$ is linear provided that $f(alpha v)=alpha f(v)$ for each scalar $alpha$ and vector $v$. Thus, if $f(v)$ were positive for some $v$ in the subspace, $f(-v)=-f(v)$ would be negative.
$endgroup$
– Brian M. Scott
Jan 9 '17 at 23:26
$begingroup$
Kill me. I facepalmed so hard. Thank you for pointing that out. The claim works out without much work. I'm used to referring that version as a "linear map" over linear function.
$endgroup$
– user64878
Jan 9 '17 at 23:30
add a comment |
3
$begingroup$
You’re using the wrong definition of linear function: in this context $f$ is linear provided that $f(alpha v)=alpha f(v)$ for each scalar $alpha$ and vector $v$. Thus, if $f(v)$ were positive for some $v$ in the subspace, $f(-v)=-f(v)$ would be negative.
$endgroup$
– Brian M. Scott
Jan 9 '17 at 23:26
$begingroup$
Kill me. I facepalmed so hard. Thank you for pointing that out. The claim works out without much work. I'm used to referring that version as a "linear map" over linear function.
$endgroup$
– user64878
Jan 9 '17 at 23:30
3
3
$begingroup$
You’re using the wrong definition of linear function: in this context $f$ is linear provided that $f(alpha v)=alpha f(v)$ for each scalar $alpha$ and vector $v$. Thus, if $f(v)$ were positive for some $v$ in the subspace, $f(-v)=-f(v)$ would be negative.
$endgroup$
– Brian M. Scott
Jan 9 '17 at 23:26
$begingroup$
You’re using the wrong definition of linear function: in this context $f$ is linear provided that $f(alpha v)=alpha f(v)$ for each scalar $alpha$ and vector $v$. Thus, if $f(v)$ were positive for some $v$ in the subspace, $f(-v)=-f(v)$ would be negative.
$endgroup$
– Brian M. Scott
Jan 9 '17 at 23:26
$begingroup$
Kill me. I facepalmed so hard. Thank you for pointing that out. The claim works out without much work. I'm used to referring that version as a "linear map" over linear function.
$endgroup$
– user64878
Jan 9 '17 at 23:30
$begingroup$
Kill me. I facepalmed so hard. Thank you for pointing that out. The claim works out without much work. I'm used to referring that version as a "linear map" over linear function.
$endgroup$
– user64878
Jan 9 '17 at 23:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Say $f(x)$ is a linear function on a subspace $mathbbS$, then
$$
forall xin mathbbS, f(x)geq 0,
$$
And
$$
-xin mathbbS, f(-x)geq 0 Longrightarrow f(-x)=-f(x)geq0,Longrightarrow f(x)leq 0
$$
Hence, $f(x)=0$.
answered Mar 25 at 8:23
ZhiyuZhiyu
111
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Say $f(x)$ is a linear function on a subspace $mathbbS$, then
$$
forall xin mathbbS, f(x)geq 0,
$$
And
$$
-xin mathbbS, f(-x)geq 0 Longrightarrow f(-x)=-f(x)geq0,Longrightarrow f(x)leq 0
$$
Hence, $f(x)=0$.
answered Mar 25 at 8:23
ZhiyuZhiyu
111
$endgroup$
add a comment |
$begingroup$
Say $f(x)$ is a linear function on a subspace $mathbbS$, then
$$
forall xin mathbbS, f(x)geq 0,
$$
And
$$
-xin mathbbS, f(-x)geq 0 Longrightarrow f(-x)=-f(x)geq0,Longrightarrow f(x)leq 0
$$
Hence, $f(x)=0$.
answered Mar 25 at 8:23
ZhiyuZhiyu
111
$endgroup$
add a comment |
$begingroup$
Say $f(x)$ is a linear function on a subspace $mathbbS$, then
$$
forall xin mathbbS, f(x)geq 0,
$$
And
$$
-xin mathbbS, f(-x)geq 0 Longrightarrow f(-x)=-f(x)geq0,Longrightarrow f(x)leq 0
$$
Hence, $f(x)=0$.
answered Mar 25 at 8:23
ZhiyuZhiyu
111
$endgroup$
Say $f(x)$ is a linear function on a subspace $mathbbS$, then
$$
forall xin mathbbS, f(x)geq 0,
$$
And
$$
-xin mathbbS, f(-x)geq 0 Longrightarrow f(-x)=-f(x)geq0,Longrightarrow f(x)leq 0
$$
Hence, $f(x)=0$.
answered Mar 25 at 8:23
ZhiyuZhiyu
111
answered Mar 25 at 8:23
ZhiyuZhiyu
111
answered Mar 25 at 8:23
ZhiyuZhiyu
111
answered Mar 25 at 8:23
ZhiyuZhiyu
111
111
add a comment |
add a comment |
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You’re using the wrong definition of linear function: in this context $f$ is linear provided that $f(alpha v)=alpha f(v)$ for each scalar $alpha$ and vector $v$. Thus, if $f(v)$ were positive for some $v$ in the subspace, $f(-v)=-f(v)$ would be negative.
$endgroup$
– Brian M. Scott
Jan 9 '17 at 23:26
$begingroup$
Kill me. I facepalmed so hard. Thank you for pointing that out. The claim works out without much work. I'm used to referring that version as a "linear map" over linear function.
$endgroup$
– user64878
Jan 9 '17 at 23:30