Volume of $(n-1)$- simplex in $n$-dimension.

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This post gives a general way to calculate $k$-simplex in $n$-dimensional space with $kleq n$. My question is, if $k=n-1$ and give vertices $v_0, cdots, v_n-1$ are linearly independent, can we show that the simplex $S$ generated by $v_0, cdots , v_n-1$ has a volume $Vol(S) = frac1n!detbeginpmatrixbf v_0 & cdots & bf v_n-1endpmatrix$?







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    Surely the volume is zero? The (affine) dimension of the set $operatornameaff v_0,...,v_n-1 $ is $n-1$. For example, if $n=2$ we can take the simplex generated by $e_1,e_2$ which is just s line segment and hence has measure zero.
    – copper.hat
    Aug 6 at 2:28











  • @copper.hat Oh, I think the case when we calculate the volume of it as that in the some $n-1$-dimensional hyperplane.
    – user124697
    Aug 6 at 2:34














up vote
0
down vote

favorite












This post gives a general way to calculate $k$-simplex in $n$-dimensional space with $kleq n$. My question is, if $k=n-1$ and give vertices $v_0, cdots, v_n-1$ are linearly independent, can we show that the simplex $S$ generated by $v_0, cdots , v_n-1$ has a volume $Vol(S) = frac1n!detbeginpmatrixbf v_0 & cdots & bf v_n-1endpmatrix$?







share|cite|improve this question















  • 1




    Surely the volume is zero? The (affine) dimension of the set $operatornameaff v_0,...,v_n-1 $ is $n-1$. For example, if $n=2$ we can take the simplex generated by $e_1,e_2$ which is just s line segment and hence has measure zero.
    – copper.hat
    Aug 6 at 2:28











  • @copper.hat Oh, I think the case when we calculate the volume of it as that in the some $n-1$-dimensional hyperplane.
    – user124697
    Aug 6 at 2:34












up vote
0
down vote

favorite









up vote
0
down vote

favorite











This post gives a general way to calculate $k$-simplex in $n$-dimensional space with $kleq n$. My question is, if $k=n-1$ and give vertices $v_0, cdots, v_n-1$ are linearly independent, can we show that the simplex $S$ generated by $v_0, cdots , v_n-1$ has a volume $Vol(S) = frac1n!detbeginpmatrixbf v_0 & cdots & bf v_n-1endpmatrix$?







share|cite|improve this question











This post gives a general way to calculate $k$-simplex in $n$-dimensional space with $kleq n$. My question is, if $k=n-1$ and give vertices $v_0, cdots, v_n-1$ are linearly independent, can we show that the simplex $S$ generated by $v_0, cdots , v_n-1$ has a volume $Vol(S) = frac1n!detbeginpmatrixbf v_0 & cdots & bf v_n-1endpmatrix$?









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 6 at 2:11









user124697

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  • 1




    Surely the volume is zero? The (affine) dimension of the set $operatornameaff v_0,...,v_n-1 $ is $n-1$. For example, if $n=2$ we can take the simplex generated by $e_1,e_2$ which is just s line segment and hence has measure zero.
    – copper.hat
    Aug 6 at 2:28











  • @copper.hat Oh, I think the case when we calculate the volume of it as that in the some $n-1$-dimensional hyperplane.
    – user124697
    Aug 6 at 2:34












  • 1




    Surely the volume is zero? The (affine) dimension of the set $operatornameaff v_0,...,v_n-1 $ is $n-1$. For example, if $n=2$ we can take the simplex generated by $e_1,e_2$ which is just s line segment and hence has measure zero.
    – copper.hat
    Aug 6 at 2:28











  • @copper.hat Oh, I think the case when we calculate the volume of it as that in the some $n-1$-dimensional hyperplane.
    – user124697
    Aug 6 at 2:34







1




1




Surely the volume is zero? The (affine) dimension of the set $operatornameaff v_0,...,v_n-1 $ is $n-1$. For example, if $n=2$ we can take the simplex generated by $e_1,e_2$ which is just s line segment and hence has measure zero.
– copper.hat
Aug 6 at 2:28





Surely the volume is zero? The (affine) dimension of the set $operatornameaff v_0,...,v_n-1 $ is $n-1$. For example, if $n=2$ we can take the simplex generated by $e_1,e_2$ which is just s line segment and hence has measure zero.
– copper.hat
Aug 6 at 2:28













@copper.hat Oh, I think the case when we calculate the volume of it as that in the some $n-1$-dimensional hyperplane.
– user124697
Aug 6 at 2:34




@copper.hat Oh, I think the case when we calculate the volume of it as that in the some $n-1$-dimensional hyperplane.
– user124697
Aug 6 at 2:34










1 Answer
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I think this is a direct consequence of the Cayley-Menger determinant.






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  • Thanks! I got it.
    – user124697
    Aug 6 at 2:35










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










I think this is a direct consequence of the Cayley-Menger determinant.






share|cite|improve this answer





















  • Thanks! I got it.
    – user124697
    Aug 6 at 2:35














up vote
1
down vote



accepted










I think this is a direct consequence of the Cayley-Menger determinant.






share|cite|improve this answer





















  • Thanks! I got it.
    – user124697
    Aug 6 at 2:35












up vote
1
down vote



accepted







up vote
1
down vote



accepted






I think this is a direct consequence of the Cayley-Menger determinant.






share|cite|improve this answer













I think this is a direct consequence of the Cayley-Menger determinant.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 6 at 2:30









Jimmy Mixco

2013




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  • Thanks! I got it.
    – user124697
    Aug 6 at 2:35
















  • Thanks! I got it.
    – user124697
    Aug 6 at 2:35















Thanks! I got it.
– user124697
Aug 6 at 2:35




Thanks! I got it.
– user124697
Aug 6 at 2:35












 

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