Identity with determinants and parallelograms

Question

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:

$\det([A,B])$ 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?

Answer 1

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 $n > 2$.

Answer 2

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 $$ \sum_{i=1}^n \det([x_1,\dots,x_{i-1},y,x_{i+1},\dots,x_n]) = \det([x_1,\dots,x_n]). $$