A sum-of-determinants identity [closed]

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.

Answer 1

I think this can also be seen pretty concretely using matrices: for clarity I'll write it out in the $n=2$ case but the general case is identical. Write the $X_j$ as column vectors and let $x_{ij}$ denote the $i^{th}$ entry of $X_j$. Then consider the product of matrices \begin{equation*} \begin{pmatrix} x_{11} & x_{12} & x_{13} \\x_{21} & x_{22} & x_{23}\\ 1 & 1 & 1\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 1\end{pmatrix} = \begin{pmatrix} x_{11}-x_{13} & x_{12}-x_{13} & x_{13} \\ x_{21}-x_{23} & x_{22}-x_{23} & x_{23} \\ 0 & 0 & 1\end{pmatrix} \end{equation*} Since the triangular matrix has determinant $1$, the two matrices with the $x$'s have equal determinants, and the two terms of the formula you want are simply the expansions of these determinants along the bottom row. (The difference in the odd/even cases is now obvious from the sign rules for expanding the determinants.)

Answer 2

Use multilinearity to expand the determinant? You'll get $$ \det(X) - \sum_{k=1}^n \det( Y(k) ) $$ where $X = [ X_1, \ldots, X_n]$ and $Y(k)$ is the matrix with the $k^\mathrm{th}$ row replaced with $X_{n+1}$. In the case $n= 2$ you get $$ \det([X_1, X_2]) - ( \det([X_1, X_3]) + \det([X_3, X_2]) = \det([X_1, X_2]) + \det([X_3, X_1]) + \det([X_2, X_3]) $$ as desired. In general, the only issue is to keep track of how many column-switches are needed to get the determinants $\det(Z(k))$ in the form you want. The sign in the formula $$ Z(k) = (-1)^t Y(k) $$ turns out to be $$ t = (n-k)+(k-1)(n-(k-1)) $$ which is $0\mod 2$ if $n$ is odd and $k\mod 2$ when $n$ is even.