Return to Question

I posed a terser version of this question on math.stackexchange.com and after 24 hours I get only a comment on a detail of notation and neither votes nor answers.

Suppose $X_1,\ldots,X_{n+1}\in\mathbb R^n$. I am guessing that \begin{align*} & \det[X_1-X_{n+1},\ldots,X_n-X_{n+1}] \\[10pt] = {} & \begin{cases} \sum_k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is even,} \\[6pt] \sum_k (-1)^k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is odd.} \end{cases} \end{align*} where addition of indices is modulo $n+1$.

My question on stackexchange was merely a reference request, but now I'm feeling I should ask whether this is known (and if so, where is it found in print)? (I think it's probably not hard to prove, so if it's not known then maybe it's only because no one happens to have been standing in the right place looking in that direction. So far all I've done is check it with random numbers for $n\le 4$.)

PS: OK, my mind was on other things than routine algebra when those other things led to this routine algebra problem. I should have shifted mental gears and thought about that.

I posed a terser version of this question on math.stackexchange.com and after 24 hours I get only a comment on a detail of notation and neither votes nor answers.

Suppose $X_1,\ldots,X_{n+1}\in\mathbb R^n$. I am guessing that \begin{align*} & \det[X_1-X_{n+1},\ldots,X_n-X_{n+1}] \\[10pt] = {} & \begin{cases} \sum_k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is even,} \\[6pt] \sum_k (-1)^k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is odd.} \end{cases} \end{align*} where addition of indices is modulo $n+1$.

My question on stackexchange was merely a reference request, but now I'm feeling I should ask whether this is known (and if so, where is it found in print)? (I think it's probably not hard to prove, so if it's not known then maybe it's only because no one happens to have been standing in the right place looking in that direction. So far all I've done is check it with random numbers for $n\le 4$.)

PS: OK, my mind was on other things than routine algebra when those other things led to this routine algebra problem. I should have shifted mental gears and thought about that.

I posed a terser version of this question on math.stackexchange.com and after 24 hours I get only a comment on a detail of notation and neither votes nor answers.

Suppose $X_1,\ldots,X_{n+1}\in\mathbb R^n$. I am guessing that \begin{align*} & \det[X_1-X_{n+1},\ldots,X_n-X_{n+1}] \\[10pt] = {} & \begin{cases} \sum_k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is even,} \\[6pt] \sum_k (-1)^k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is odd.} \end{cases} \end{align*} where addition of indices is modulo $n+1$.

My question on stackexchange was merely a reference request, but now I'm feeling I should ask whether this is known (and if so, where is it found in print)? (I think it's probably not hard to prove, so if it's not known then maybe it's only because no one happens to have been standing in the right place looking in that direction. So far all I've done is check it with random numbers for $n\le 4$.)

I posed a terser version of this question on math.stackexchange.com and after 24 hours I get only a comment on a detail of notation and neither votes nor answers.

Suppose $X_1,\ldots,X_{n+1}\in\mathbb R^n$. I am guessing that \begin{align*} & \det[X_1-X_{n+1},\ldots,X_n-X_{n+1}] \\[10pt] = {} & \begin{cases} \sum_k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is even,} \\[6pt] \sum_k (-1)^k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is odd.} \end{cases} \end{align*} where addition of indices is modulo $n+1$.

My question on stackexchange was merely a reference request, but now I'm feeling I should ask whether this is known (and if so, where is it found in print)? (I think it's probably not hard to prove, so if it's not known then maybe it's only because no one happens to have been standing in the right place looking in that direction. So far all I've done is check it with random numbers for $n\le 4$.)

PS: OK, my mind was on other things than routine algebra when those other things led to this routine algebra problem. I should have shifted mental gears and thought about that.

I posed a terser version of this question on math.stackexchange.com and after 24 hours I get only a comment on a detail of notation and neither votes nor answers.

Suppose $X_1,\ldots,X_{n+1}\in\mathbb R^n$. I am guessing that \begin{align*} & \det[X_1-X_{n+1},\ldots,X_n-X_{n+1}] \\[10pt] = {} & \begin{cases} \sum_k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is odd}, \\[6pt] \sum_k (-1)^k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is even.} \end{cases} \end{align*} where addition of indices is modulo $n$.

My question on stackexchange was merely a reference request, but now I'm feeling I should ask whether this is known (and if so, where is it found in print)? (I think it's probably not hard to prove, so if it's not known then maybe it's only because no one happens to have been standing in the right place looking in that direction. So far all I've done is check it with random numbers for $n\le 4$.)

I posed a terser version of this question on math.stackexchange.com and after 24 hours I get only a comment on a detail of notation and neither votes nor answers.

Suppose $X_1,\ldots,X_{n+1}\in\mathbb R^n$. I am guessing that \begin{align*} & \det[X_1-X_{n+1},\ldots,X_n-X_{n+1}] \\[10pt] = {} & \begin{cases} \sum_k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is even,} \\[6pt] \sum_k (-1)^k\det[X_{k+1},\ldots,X_{k+n}] & \text{if $n$ is odd.} \end{cases} \end{align*} where addition of indices is modulo $n+1$.

My question on stackexchange was merely a reference request, but now I'm feeling I should ask whether this is known (and if so, where is it found in print)? (I think it's probably not hard to prove, so if it's not known then maybe it's only because no one happens to have been standing in the right place looking in that direction. So far all I've done is check it with random numbers for $n\le 4$.)