Understand a definition
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$begingroup$
Reading Randomized Rounding without Solving the Linear Program paper I came across this definition:
Let $P$ be a convex set in $mathbb{R}^n$ and let $f$ be a linear function (not necessarily homogeneous) from $P$ to $mathbb{R}^m$.
Correct me if I am wrong, but $f$ is a non-homogenous function that takes a set of points in $mathbb{R}^n$ and returns a set of points in $mathbb{R}^m$?
linear-algebra functions discrete-mathematics
edited Jan 24 at 21:13
gt6989b
34.7k22456
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
$endgroup$
add a comment |
$begingroup$
Reading Randomized Rounding without Solving the Linear Program paper I came across this definition:
Let $P$ be a convex set in $mathbb{R}^n$ and let $f$ be a linear function (not necessarily homogeneous) from $P$ to $mathbb{R}^m$.
Correct me if I am wrong, but $f$ is a non-homogenous function that takes a set of points in $mathbb{R}^n$ and returns a set of points in $mathbb{R}^m$?
linear-algebra functions discrete-mathematics
edited Jan 24 at 21:13
gt6989b
34.7k22456
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
$endgroup$
add a comment |
$begingroup$
Reading Randomized Rounding without Solving the Linear Program paper I came across this definition:
Let $P$ be a convex set in $mathbb{R}^n$ and let $f$ be a linear function (not necessarily homogeneous) from $P$ to $mathbb{R}^m$.
Correct me if I am wrong, but $f$ is a non-homogenous function that takes a set of points in $mathbb{R}^n$ and returns a set of points in $mathbb{R}^m$?
linear-algebra functions discrete-mathematics
edited Jan 24 at 21:13
gt6989b
34.7k22456
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
$endgroup$
Reading Randomized Rounding without Solving the Linear Program paper I came across this definition:
Let $P$ be a convex set in $mathbb{R}^n$ and let $f$ be a linear function (not necessarily homogeneous) from $P$ to $mathbb{R}^m$.
Correct me if I am wrong, but $f$ is a non-homogenous function that takes a set of points in $mathbb{R}^n$ and returns a set of points in $mathbb{R}^m$?
linear-algebra functions discrete-mathematics
linear-algebra functions discrete-mathematics
edited Jan 24 at 21:13
gt6989b
34.7k22456
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
edited Jan 24 at 21:13
gt6989b
34.7k22456
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
edited Jan 24 at 21:13
gt6989b
34.7k22456
edited Jan 24 at 21:13
gt6989b
34.7k22456
edited Jan 24 at 21:13
gt6989b
34.7k22456
34.7k22456
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
asked Jan 24 at 21:10
Tony TannousTony Tannous
1676
1676
add a comment |
add a comment |
1 Answer
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$begingroup$
A homogeneous function $f$ satisfies $f(ax)=af(x)$ for scalars.
When $f(x)=Ax+b$ is linear, this is to say that $b=0$. So a non-homogeneous linear function may have a non-zero constant vector term.
So in your case, $f$ may or may not be homogeneous/non-homogeneous. Strictly speaking it doesn't map $mathbb{R}^ntomathbb{R}^m$, it maps $Ptomathbb{R}^m$. But since $f$ is linear, the extension is a natural one, and it makes good sense to just think of it as mapping $mathbb{R}^ntomathbb{R}^m$.
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
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add a comment |
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1 Answer
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$begingroup$
A homogeneous function $f$ satisfies $f(ax)=af(x)$ for scalars.
When $f(x)=Ax+b$ is linear, this is to say that $b=0$. So a non-homogeneous linear function may have a non-zero constant vector term.
So in your case, $f$ may or may not be homogeneous/non-homogeneous. Strictly speaking it doesn't map $mathbb{R}^ntomathbb{R}^m$, it maps $Ptomathbb{R}^m$. But since $f$ is linear, the extension is a natural one, and it makes good sense to just think of it as mapping $mathbb{R}^ntomathbb{R}^m$.
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
add a comment |
$begingroup$
A homogeneous function $f$ satisfies $f(ax)=af(x)$ for scalars.
When $f(x)=Ax+b$ is linear, this is to say that $b=0$. So a non-homogeneous linear function may have a non-zero constant vector term.
So in your case, $f$ may or may not be homogeneous/non-homogeneous. Strictly speaking it doesn't map $mathbb{R}^ntomathbb{R}^m$, it maps $Ptomathbb{R}^m$. But since $f$ is linear, the extension is a natural one, and it makes good sense to just think of it as mapping $mathbb{R}^ntomathbb{R}^m$.
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
add a comment |
$begingroup$
A homogeneous function $f$ satisfies $f(ax)=af(x)$ for scalars.
When $f(x)=Ax+b$ is linear, this is to say that $b=0$. So a non-homogeneous linear function may have a non-zero constant vector term.
So in your case, $f$ may or may not be homogeneous/non-homogeneous. Strictly speaking it doesn't map $mathbb{R}^ntomathbb{R}^m$, it maps $Ptomathbb{R}^m$. But since $f$ is linear, the extension is a natural one, and it makes good sense to just think of it as mapping $mathbb{R}^ntomathbb{R}^m$.
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
A homogeneous function $f$ satisfies $f(ax)=af(x)$ for scalars.
When $f(x)=Ax+b$ is linear, this is to say that $b=0$. So a non-homogeneous linear function may have a non-zero constant vector term.
So in your case, $f$ may or may not be homogeneous/non-homogeneous. Strictly speaking it doesn't map $mathbb{R}^ntomathbb{R}^m$, it maps $Ptomathbb{R}^m$. But since $f$ is linear, the extension is a natural one, and it makes good sense to just think of it as mapping $mathbb{R}^ntomathbb{R}^m$.
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
edited Jan 24 at 21:22
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
answered Jan 24 at 21:14
nathan.j.mcdougallnathan.j.mcdougall
1,519818
1,519818
add a comment |
add a comment |
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