Prove that 0 is in the convex hull of points chosen from each orthant
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If we arbitrarily choose a point from each orthant in $mathbbR^n$, that is we choose $2^n$ points in total, how do we prove that 0 is in the convex hull of these $2^n$ points? It seems obvious, but when I sit down and start thinking about it, I couldn't definitely find a set of $lambda_i, i = 1,2,cdots,2^n$ such that $0 leq lambda_i leq 1, sum_i lambda_i = 1$ and $0 = sum_i lambda_i x_i$, where $x_i, i = 1,2,cdots,2^n$ are any set of points satisfying the condition.
convex-analysis convex-geometry convex-hulls
asked Jan 28 at 18:03
Sean IanSean Ian
1035
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add a comment |
$begingroup$
If we arbitrarily choose a point from each orthant in $mathbbR^n$, that is we choose $2^n$ points in total, how do we prove that 0 is in the convex hull of these $2^n$ points? It seems obvious, but when I sit down and start thinking about it, I couldn't definitely find a set of $lambda_i, i = 1,2,cdots,2^n$ such that $0 leq lambda_i leq 1, sum_i lambda_i = 1$ and $0 = sum_i lambda_i x_i$, where $x_i, i = 1,2,cdots,2^n$ are any set of points satisfying the condition.
convex-analysis convex-geometry convex-hulls
asked Jan 28 at 18:03
Sean IanSean Ian
1035
$endgroup$
add a comment |
$begingroup$
If we arbitrarily choose a point from each orthant in $mathbbR^n$, that is we choose $2^n$ points in total, how do we prove that 0 is in the convex hull of these $2^n$ points? It seems obvious, but when I sit down and start thinking about it, I couldn't definitely find a set of $lambda_i, i = 1,2,cdots,2^n$ such that $0 leq lambda_i leq 1, sum_i lambda_i = 1$ and $0 = sum_i lambda_i x_i$, where $x_i, i = 1,2,cdots,2^n$ are any set of points satisfying the condition.
convex-analysis convex-geometry convex-hulls
asked Jan 28 at 18:03
Sean IanSean Ian
1035
$endgroup$
If we arbitrarily choose a point from each orthant in $mathbbR^n$, that is we choose $2^n$ points in total, how do we prove that 0 is in the convex hull of these $2^n$ points? It seems obvious, but when I sit down and start thinking about it, I couldn't definitely find a set of $lambda_i, i = 1,2,cdots,2^n$ such that $0 leq lambda_i leq 1, sum_i lambda_i = 1$ and $0 = sum_i lambda_i x_i$, where $x_i, i = 1,2,cdots,2^n$ are any set of points satisfying the condition.
convex-analysis convex-geometry convex-hulls
convex-analysis convex-geometry convex-hulls
asked Jan 28 at 18:03
Sean IanSean Ian
1035
asked Jan 28 at 18:03
Sean IanSean Ian
1035
asked Jan 28 at 18:03
Sean IanSean Ian
1035
asked Jan 28 at 18:03
Sean IanSean Ian
1035
asked Jan 28 at 18:03
Sean IanSean Ian
1035
1035
add a comment |
add a comment |
2 Answers
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$begingroup$
Induct on dimension.
In the $n=1$ base case, you have a positive number and a negative number; $0$ can be represented as a convex combination of them.
For $n>1$: Given your $2^n$ points, one from each orthant, divide them into 2 sets of size $2^n-1$, where the first set has points whose last coordinates are positive and the second set where the last coordinates are negative. By the inductive hypothesis there is a convex combo of the first set of points such that the first $n-1$ coordinates vanish, and similarly for the second set. The last coordinates of these two convex combos are positive and negative, so there is a convex combo of them that is zero.
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
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$begingroup$
Take in pairs the points that differ in sign for the coordinate $n$ (and only that one) and form the convex combinations such that that coordinate is cancelled.
Now you have $2^n-1$ points in each orthants of the diminished space.
The results holds because the convex combination of linear combinations is a convex combination.
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
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2 Answers
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active
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2 Answers
2
active
oldest
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votes
$begingroup$
Induct on dimension.
In the $n=1$ base case, you have a positive number and a negative number; $0$ can be represented as a convex combination of them.
For $n>1$: Given your $2^n$ points, one from each orthant, divide them into 2 sets of size $2^n-1$, where the first set has points whose last coordinates are positive and the second set where the last coordinates are negative. By the inductive hypothesis there is a convex combo of the first set of points such that the first $n-1$ coordinates vanish, and similarly for the second set. The last coordinates of these two convex combos are positive and negative, so there is a convex combo of them that is zero.
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
$endgroup$
add a comment |
$begingroup$
Induct on dimension.
In the $n=1$ base case, you have a positive number and a negative number; $0$ can be represented as a convex combination of them.
For $n>1$: Given your $2^n$ points, one from each orthant, divide them into 2 sets of size $2^n-1$, where the first set has points whose last coordinates are positive and the second set where the last coordinates are negative. By the inductive hypothesis there is a convex combo of the first set of points such that the first $n-1$ coordinates vanish, and similarly for the second set. The last coordinates of these two convex combos are positive and negative, so there is a convex combo of them that is zero.
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
$endgroup$
add a comment |
$begingroup$
Induct on dimension.
In the $n=1$ base case, you have a positive number and a negative number; $0$ can be represented as a convex combination of them.
For $n>1$: Given your $2^n$ points, one from each orthant, divide them into 2 sets of size $2^n-1$, where the first set has points whose last coordinates are positive and the second set where the last coordinates are negative. By the inductive hypothesis there is a convex combo of the first set of points such that the first $n-1$ coordinates vanish, and similarly for the second set. The last coordinates of these two convex combos are positive and negative, so there is a convex combo of them that is zero.
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
$endgroup$
Induct on dimension.
In the $n=1$ base case, you have a positive number and a negative number; $0$ can be represented as a convex combination of them.
For $n>1$: Given your $2^n$ points, one from each orthant, divide them into 2 sets of size $2^n-1$, where the first set has points whose last coordinates are positive and the second set where the last coordinates are negative. By the inductive hypothesis there is a convex combo of the first set of points such that the first $n-1$ coordinates vanish, and similarly for the second set. The last coordinates of these two convex combos are positive and negative, so there is a convex combo of them that is zero.
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
edited Jan 28 at 18:53
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
answered Jan 28 at 18:20
kimchi loverkimchi lover
10.7k31128
10.7k31128
add a comment |
add a comment |
$begingroup$
Take in pairs the points that differ in sign for the coordinate $n$ (and only that one) and form the convex combinations such that that coordinate is cancelled.
Now you have $2^n-1$ points in each orthants of the diminished space.
The results holds because the convex combination of linear combinations is a convex combination.
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Take in pairs the points that differ in sign for the coordinate $n$ (and only that one) and form the convex combinations such that that coordinate is cancelled.
Now you have $2^n-1$ points in each orthants of the diminished space.
The results holds because the convex combination of linear combinations is a convex combination.
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Take in pairs the points that differ in sign for the coordinate $n$ (and only that one) and form the convex combinations such that that coordinate is cancelled.
Now you have $2^n-1$ points in each orthants of the diminished space.
The results holds because the convex combination of linear combinations is a convex combination.
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
$endgroup$
Take in pairs the points that differ in sign for the coordinate $n$ (and only that one) and form the convex combinations such that that coordinate is cancelled.
Now you have $2^n-1$ points in each orthants of the diminished space.
The results holds because the convex combination of linear combinations is a convex combination.
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
edited Jan 28 at 18:53
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
answered Jan 28 at 18:46
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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