standard n-simplex
$begingroup$
We know that a n-simplex is a convex hull of n+1 affinely independents points in $mathbb{R}^n$, i,e, let $ x_0,x_1, ldots ,x_n $ affinely independents points in $mathbb{R}^n$ then the n-simplex determinated is :
$$ C=lbrace lambda_0 x_0 + lambda_1 x_1+ ldots + lambda_n x_n / lambda_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace $$
Also a standard n-simplex, $Delta^n$, is the set :
$$ Delta^n=lbrace (t_0,t_1,ldots,t_n) in mathbb{R}^{n+1}/t_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace .$$
It is noted that the standard n-simplex is the convex hull of the canonical vectors $e_0,e_1, ldots, e_n in mathbb{R}^{n+1}$
We also see that the set $X=lbrace 0,e_0,e_1,ldots,e_n rbrace$ is affinely independent in $mathbb{R}^{n+1}$ so following the definition of n+1-simplex we should have that $Delta^n$ is a $n+1$-simplex because the convex hull of $X$ is $Delta^n$.
Why do the n-simplex standard define it that way if it is really a n + 1 simplex?
simplex
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
$endgroup$
add a comment |
$begingroup$
We know that a n-simplex is a convex hull of n+1 affinely independents points in $mathbb{R}^n$, i,e, let $ x_0,x_1, ldots ,x_n $ affinely independents points in $mathbb{R}^n$ then the n-simplex determinated is :
$$ C=lbrace lambda_0 x_0 + lambda_1 x_1+ ldots + lambda_n x_n / lambda_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace $$
Also a standard n-simplex, $Delta^n$, is the set :
$$ Delta^n=lbrace (t_0,t_1,ldots,t_n) in mathbb{R}^{n+1}/t_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace .$$
It is noted that the standard n-simplex is the convex hull of the canonical vectors $e_0,e_1, ldots, e_n in mathbb{R}^{n+1}$
We also see that the set $X=lbrace 0,e_0,e_1,ldots,e_n rbrace$ is affinely independent in $mathbb{R}^{n+1}$ so following the definition of n+1-simplex we should have that $Delta^n$ is a $n+1$-simplex because the convex hull of $X$ is $Delta^n$.
Why do the n-simplex standard define it that way if it is really a n + 1 simplex?
simplex
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
$endgroup$
add a comment |
$begingroup$
We know that a n-simplex is a convex hull of n+1 affinely independents points in $mathbb{R}^n$, i,e, let $ x_0,x_1, ldots ,x_n $ affinely independents points in $mathbb{R}^n$ then the n-simplex determinated is :
$$ C=lbrace lambda_0 x_0 + lambda_1 x_1+ ldots + lambda_n x_n / lambda_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace $$
Also a standard n-simplex, $Delta^n$, is the set :
$$ Delta^n=lbrace (t_0,t_1,ldots,t_n) in mathbb{R}^{n+1}/t_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace .$$
It is noted that the standard n-simplex is the convex hull of the canonical vectors $e_0,e_1, ldots, e_n in mathbb{R}^{n+1}$
We also see that the set $X=lbrace 0,e_0,e_1,ldots,e_n rbrace$ is affinely independent in $mathbb{R}^{n+1}$ so following the definition of n+1-simplex we should have that $Delta^n$ is a $n+1$-simplex because the convex hull of $X$ is $Delta^n$.
Why do the n-simplex standard define it that way if it is really a n + 1 simplex?
simplex
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
$endgroup$
We know that a n-simplex is a convex hull of n+1 affinely independents points in $mathbb{R}^n$, i,e, let $ x_0,x_1, ldots ,x_n $ affinely independents points in $mathbb{R}^n$ then the n-simplex determinated is :
$$ C=lbrace lambda_0 x_0 + lambda_1 x_1+ ldots + lambda_n x_n / lambda_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace $$
Also a standard n-simplex, $Delta^n$, is the set :
$$ Delta^n=lbrace (t_0,t_1,ldots,t_n) in mathbb{R}^{n+1}/t_i geq 0, sum_{i=0}^{n}lambda_i =1 rbrace .$$
It is noted that the standard n-simplex is the convex hull of the canonical vectors $e_0,e_1, ldots, e_n in mathbb{R}^{n+1}$
We also see that the set $X=lbrace 0,e_0,e_1,ldots,e_n rbrace$ is affinely independent in $mathbb{R}^{n+1}$ so following the definition of n+1-simplex we should have that $Delta^n$ is a $n+1$-simplex because the convex hull of $X$ is $Delta^n$.
Why do the n-simplex standard define it that way if it is really a n + 1 simplex?
simplex
simplex
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
edited Dec 4 '18 at 1:37
Ethan Bolker
42.2k548111
42.2k548111
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
asked Dec 4 '18 at 1:32
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
154
154
add a comment |
add a comment |
1 Answer
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The convex hull of $X$ is an ($n + 1$)-simplex, that's true, but it's not the same as $Delta^n$. For example, take $n = 1$. $Delta^n$ is the line segment from $(0, 1)$ to $(1, 0)$, but the convex hull of $X = {(0, 0), (1, 0), (0, 1)}$ is a solid triangle.
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
$endgroup$
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The convex hull of $X$ is an ($n + 1$)-simplex, that's true, but it's not the same as $Delta^n$. For example, take $n = 1$. $Delta^n$ is the line segment from $(0, 1)$ to $(1, 0)$, but the convex hull of $X = {(0, 0), (1, 0), (0, 1)}$ is a solid triangle.
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
The convex hull of $X$ is an ($n + 1$)-simplex, that's true, but it's not the same as $Delta^n$. For example, take $n = 1$. $Delta^n$ is the line segment from $(0, 1)$ to $(1, 0)$, but the convex hull of $X = {(0, 0), (1, 0), (0, 1)}$ is a solid triangle.
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
The convex hull of $X$ is an ($n + 1$)-simplex, that's true, but it's not the same as $Delta^n$. For example, take $n = 1$. $Delta^n$ is the line segment from $(0, 1)$ to $(1, 0)$, but the convex hull of $X = {(0, 0), (1, 0), (0, 1)}$ is a solid triangle.
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
$endgroup$
The convex hull of $X$ is an ($n + 1$)-simplex, that's true, but it's not the same as $Delta^n$. For example, take $n = 1$. $Delta^n$ is the line segment from $(0, 1)$ to $(1, 0)$, but the convex hull of $X = {(0, 0), (1, 0), (0, 1)}$ is a solid triangle.
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
edited Dec 4 '18 at 3:32
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
answered Dec 4 '18 at 3:23
Hew WolffHew Wolff
2,245716
2,245716
add a comment |
add a comment |
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