Mixing
The unit cube is the convex combination of its vertices
$begingroup$
I want to prove this statement:
$[0,1]^d = conv(v_1,...,v_n)$, with $n = 2^d$ and ${v_1,...,v_n} = {0,1}^d$.
Geometrically this is evident, I'm looking for a pure algebraic proof.
convex-analysis
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
$endgroup$
add a comment |
$begingroup$
I want to prove this statement:
$[0,1]^d = conv(v_1,...,v_n)$, with $n = 2^d$ and ${v_1,...,v_n} = {0,1}^d$.
Geometrically this is evident, I'm looking for a pure algebraic proof.
convex-analysis
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
$endgroup$
$begingroup$
I think you need induction.
$endgroup$
– Wuestenfux
Jan 1 at 10:50
$begingroup$
@Wuestenfux why? it's just directly obvious.
$endgroup$
– mathworker21
Jan 1 at 10:56
add a comment |
$begingroup$
I want to prove this statement:
$[0,1]^d = conv(v_1,...,v_n)$, with $n = 2^d$ and ${v_1,...,v_n} = {0,1}^d$.
Geometrically this is evident, I'm looking for a pure algebraic proof.
convex-analysis
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
$endgroup$
I want to prove this statement:
$[0,1]^d = conv(v_1,...,v_n)$, with $n = 2^d$ and ${v_1,...,v_n} = {0,1}^d$.
Geometrically this is evident, I'm looking for a pure algebraic proof.
convex-analysis
convex-analysis
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
asked Jan 1 at 10:47
Markus SteinerMarkus Steiner
785
785
$begingroup$
I think you need induction.
$endgroup$
– Wuestenfux
Jan 1 at 10:50
$begingroup$
@Wuestenfux why? it's just directly obvious.
$endgroup$
– mathworker21
Jan 1 at 10:56
add a comment |
$begingroup$
I think you need induction.
$endgroup$
– Wuestenfux
Jan 1 at 10:50
$begingroup$
@Wuestenfux why? it's just directly obvious.
$endgroup$
– mathworker21
Jan 1 at 10:56
$begingroup$
I think you need induction.
$endgroup$
– Wuestenfux
Jan 1 at 10:50
$begingroup$
I think you need induction.
$endgroup$
– Wuestenfux
Jan 1 at 10:50
$begingroup$
@Wuestenfux why? it's just directly obvious.
$endgroup$
– mathworker21
Jan 1 at 10:56
$begingroup$
@Wuestenfux why? it's just directly obvious.
$endgroup$
– mathworker21
Jan 1 at 10:56
add a comment |
1 Answer
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All right. In two parts:
First, that all convex combinations are contained in the cube. For each $i$, the $i$th coordinate $x_i$ is a convex combination of the $i$th coordinates of the vertices $v_j$, each of which is $0$ or $1$. Such a convex combination must lie in $[0,1]$. Repeat over all $i$, and the point lies in the unit cube ${x: x_iin [0,1]text{ for all }i}$.
Second, that all points in the cube are convex combinations. For this, we just find a convex combination explicitly. Index the vertices with subsets of ${1,2,dots,d}$; the vertex $v_S$ has $x_i=1$ if $iin S$ and $x_i=0$ otherwise.
Now, to reach a point $u_i$ with coordinates $u_i$, we give it coefficients as follows:
$$u = sum_{S} left(prod_{iin S}u_iright)left(prod_{inotin S}1-u_iright)v_S$$
Now, in order to extract a particular coordinate $u_k$, sort the $v_S$ by their $k$th coordinate. That coordinate is $1$ if $kin S$ and $0$ otherwise, so the nonzero terms always give us $u_k$ rather than $1-u_k$ in that product. Every other term in the product freely varies between the two possibilities, covering all $2^{d-1}$ options - and the sum of those products is therefore $1cdot 1cdots 1cdot u_kcdot 1cdots = u_k$. Done.
answered Jan 1 at 11:32
jmerryjmerry
3,054412
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add a comment |
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1 Answer
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$begingroup$
All right. In two parts:
First, that all convex combinations are contained in the cube. For each $i$, the $i$th coordinate $x_i$ is a convex combination of the $i$th coordinates of the vertices $v_j$, each of which is $0$ or $1$. Such a convex combination must lie in $[0,1]$. Repeat over all $i$, and the point lies in the unit cube ${x: x_iin [0,1]text{ for all }i}$.
Second, that all points in the cube are convex combinations. For this, we just find a convex combination explicitly. Index the vertices with subsets of ${1,2,dots,d}$; the vertex $v_S$ has $x_i=1$ if $iin S$ and $x_i=0$ otherwise.
Now, to reach a point $u_i$ with coordinates $u_i$, we give it coefficients as follows:
$$u = sum_{S} left(prod_{iin S}u_iright)left(prod_{inotin S}1-u_iright)v_S$$
Now, in order to extract a particular coordinate $u_k$, sort the $v_S$ by their $k$th coordinate. That coordinate is $1$ if $kin S$ and $0$ otherwise, so the nonzero terms always give us $u_k$ rather than $1-u_k$ in that product. Every other term in the product freely varies between the two possibilities, covering all $2^{d-1}$ options - and the sum of those products is therefore $1cdot 1cdots 1cdot u_kcdot 1cdots = u_k$. Done.
answered Jan 1 at 11:32
jmerryjmerry
3,054412
$endgroup$
add a comment |
$begingroup$
All right. In two parts:
First, that all convex combinations are contained in the cube. For each $i$, the $i$th coordinate $x_i$ is a convex combination of the $i$th coordinates of the vertices $v_j$, each of which is $0$ or $1$. Such a convex combination must lie in $[0,1]$. Repeat over all $i$, and the point lies in the unit cube ${x: x_iin [0,1]text{ for all }i}$.
Second, that all points in the cube are convex combinations. For this, we just find a convex combination explicitly. Index the vertices with subsets of ${1,2,dots,d}$; the vertex $v_S$ has $x_i=1$ if $iin S$ and $x_i=0$ otherwise.
Now, to reach a point $u_i$ with coordinates $u_i$, we give it coefficients as follows:
$$u = sum_{S} left(prod_{iin S}u_iright)left(prod_{inotin S}1-u_iright)v_S$$
Now, in order to extract a particular coordinate $u_k$, sort the $v_S$ by their $k$th coordinate. That coordinate is $1$ if $kin S$ and $0$ otherwise, so the nonzero terms always give us $u_k$ rather than $1-u_k$ in that product. Every other term in the product freely varies between the two possibilities, covering all $2^{d-1}$ options - and the sum of those products is therefore $1cdot 1cdots 1cdot u_kcdot 1cdots = u_k$. Done.
answered Jan 1 at 11:32
jmerryjmerry
3,054412
$endgroup$
add a comment |
$begingroup$
All right. In two parts:
First, that all convex combinations are contained in the cube. For each $i$, the $i$th coordinate $x_i$ is a convex combination of the $i$th coordinates of the vertices $v_j$, each of which is $0$ or $1$. Such a convex combination must lie in $[0,1]$. Repeat over all $i$, and the point lies in the unit cube ${x: x_iin [0,1]text{ for all }i}$.
Second, that all points in the cube are convex combinations. For this, we just find a convex combination explicitly. Index the vertices with subsets of ${1,2,dots,d}$; the vertex $v_S$ has $x_i=1$ if $iin S$ and $x_i=0$ otherwise.
Now, to reach a point $u_i$ with coordinates $u_i$, we give it coefficients as follows:
$$u = sum_{S} left(prod_{iin S}u_iright)left(prod_{inotin S}1-u_iright)v_S$$
Now, in order to extract a particular coordinate $u_k$, sort the $v_S$ by their $k$th coordinate. That coordinate is $1$ if $kin S$ and $0$ otherwise, so the nonzero terms always give us $u_k$ rather than $1-u_k$ in that product. Every other term in the product freely varies between the two possibilities, covering all $2^{d-1}$ options - and the sum of those products is therefore $1cdot 1cdots 1cdot u_kcdot 1cdots = u_k$. Done.
answered Jan 1 at 11:32
jmerryjmerry
3,054412
$endgroup$
All right. In two parts:
First, that all convex combinations are contained in the cube. For each $i$, the $i$th coordinate $x_i$ is a convex combination of the $i$th coordinates of the vertices $v_j$, each of which is $0$ or $1$. Such a convex combination must lie in $[0,1]$. Repeat over all $i$, and the point lies in the unit cube ${x: x_iin [0,1]text{ for all }i}$.
Second, that all points in the cube are convex combinations. For this, we just find a convex combination explicitly. Index the vertices with subsets of ${1,2,dots,d}$; the vertex $v_S$ has $x_i=1$ if $iin S$ and $x_i=0$ otherwise.
Now, to reach a point $u_i$ with coordinates $u_i$, we give it coefficients as follows:
$$u = sum_{S} left(prod_{iin S}u_iright)left(prod_{inotin S}1-u_iright)v_S$$
Now, in order to extract a particular coordinate $u_k$, sort the $v_S$ by their $k$th coordinate. That coordinate is $1$ if $kin S$ and $0$ otherwise, so the nonzero terms always give us $u_k$ rather than $1-u_k$ in that product. Every other term in the product freely varies between the two possibilities, covering all $2^{d-1}$ options - and the sum of those products is therefore $1cdot 1cdots 1cdot u_kcdot 1cdots = u_k$. Done.
answered Jan 1 at 11:32
jmerryjmerry
3,054412
answered Jan 1 at 11:32
jmerryjmerry
3,054412
answered Jan 1 at 11:32
jmerryjmerry
3,054412
answered Jan 1 at 11:32
jmerryjmerry
3,054412
3,054412
add a comment |
add a comment |
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$begingroup$
I think you need induction.
$endgroup$
– Wuestenfux
Jan 1 at 10:50
$begingroup$
@Wuestenfux why? it's just directly obvious.
$endgroup$
– mathworker21
Jan 1 at 10:56