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While working on determinants, I found the following identity. Given two vectors $A,B \in \mathbb{R}^2$, let $\det([A,B])$ be the determinant of the matrix whose columns are $A$ and $B$. If $B$ is any vector on the segment $AC$, then:

$$\det([A,B])+\det([B,C]) = \det([A,C])$$

The proof is by picture:

enter image description here

$\det([A,B])$ is the signed area of the parallelogram determined by the vectors $A,B$. Similarly, $\det([B,C])$ is the signed area of the parallelogram determined by $B,C$. It is easy to see that the sum of these areas equals the signed area of the parallelogram determined by $A,C$.

MY QUESTION IS: what is the generalization of this identity to $n$ dimensions?

Erel Segal-Halevi
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2 Answers2

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Given $n$ column vectors $x_1,\dots,x_n\in\mathbb R^n$ and a vector $y$ in the convex (or even affine) hull of $x_1,\dots,x_n$, then $$ \sum_{i=1}^n \det([x_1,\dots,x_{i-1},y,x_{i+1},\dots,x_n]) = \det([x_1,\dots,x_n]). $$

Christoph
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Your equation can be extended to any two $n$-dimensional columns $A,C$ in an $n\times n$-determinant:

For $n=2$ We have $B = pA+(1-p)C$ for $p \in (0,1)$:

$$\det([A,B])+\det([B,C]) = \det([A,pA+(1-p)C])+\det([pA+(1-p)C,C]) = $$ $$ =(1-p)\det([A,C]) + p\det([A,C]) = \det([A,C])$$

The calculation remains the same for $n > 2$.

trancelocation
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  • What is the determinant of $[A,C]$ when $n>2$? – Erel Segal-Halevi Apr 09 '18 at 13:20
  • For $n>2$ you just have more columns ($c_1 \ldots c_n$) and then $A$ and $C$ are two of those $n$-dimensnional columns. $[c_1\ldots c_i, \ldots c_j,\ldots , c_n] =[c_1\ldots A, \ldots C,\ldots , c_n]$. – trancelocation Apr 09 '18 at 13:26
  • So for example, with $n=5$ we will have: $\det([X,A,Y,B,Z]) + \det([X,B,Y,C,Z]) = \det([X,A,Y,C,Z])$? – Erel Segal-Halevi Apr 09 '18 at 14:08
  • Yeah! Exactly! This is what the calculation actually shows, because the determinant is linear in every column. The only difference is that $A$ and $C$ are now columns with 5 entries each. – trancelocation Apr 09 '18 at 14:10
  • Now I see! It is even neater than my geometric proof. And, $p$ can be anything - not necessarily in $(0,1)$. – Erel Segal-Halevi Apr 09 '18 at 14:40
  • I'd argue that this variant is just the two-dimensional version with an appendix. See my answer for a more general formula. – Christoph Apr 10 '18 at 07:45