-1
$\begingroup$

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.

$\endgroup$
6
  • 5
    $\begingroup$ Doesn't this follow easily from the multilinearity of the determinant? $\endgroup$ Dec 10, 2014 at 19:48
  • $\begingroup$ @RichardStanley : Could be --- I was thinking about geometry, not algebra. $\endgroup$ Dec 10, 2014 at 19:49
  • $\begingroup$ $\ldots$ which just goes to show that algebra is efficacious; I should have shifted mental gears and thought of using it $\ldots$ $\endgroup$ Dec 10, 2014 at 19:50
  • 6
    $\begingroup$ You are computing the determinant of a matrix $[X_1,\dots,X_n]$ updated by a rank one matrix $X_n (-1,\dots,-1)$, and the matrix determinant lemma followed by Cramer's rule gives the claim. en.wikipedia.org/wiki/Matrix_determinant_lemma en.wikipedia.org/wiki/Cramer%27s_rule $\endgroup$
    – Terry Tao
    Dec 11, 2014 at 2:52
  • 1
    $\begingroup$ It's disappointing that you didn't get a quick answer on math.stackexchange.com. $\endgroup$
    – Deane Yang
    Dec 11, 2014 at 3:36

2 Answers 2

2
$\begingroup$

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.)

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ I should have torn my brain away from the geometry question I was thinking about and thought: obviously this is algebra. Wondering about the geometric meaning in higher dimensions, of the choice of $\pm$ in each term was actually the main thing on my mind, and I haven't really done much with that yet. In dimension $2$ there are cases where one obviously wants to subtract, rather than add, the area of a triangle, and that's taken care of by the fact that the determinant is negative in those cases. $\endgroup$ Dec 10, 2014 at 20:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.