# A combinatorial proof of determinant as the hyper-volume bounded by vectors?

There are many proofs that the determinant of a 2x2 matrix is $$ad - bc$$ which is the area of a parallelogram bounded by the row (or column) vectors of the matrix. They come in many forms: plain calculation, induction, proof by picture or geometry, non-computational properties of a determinant and the area. Some of these translate to higher dimensions (a simple search on google/math.SE shows many variations of these).

The determinant for $$n$$ matrices is often given as a direct computation over permutations:

$$\det A = \sum_{\pi \in S_n} sgn(\pi) \cdot a_{1\pi(1)} \cdot \ldots \cdot a_{n\pi(n)}$$ where $$S_n = \{ \pi : \{1,\ldots, n\} \to \{1,\ldots, n\} ~ | ~ \pi \text{ bijective }\}$$ and $$sgn(\pi)$$ the is the parity of the permutation $$\pi$$, i.e. either $$1$$ or $$-1$$.

I wonder if there is an illuminating proof for $$n$$ dimensions that uses permutations combinatorially, an extension of the proof by skewing areas and volumes and adding or subtracting areas/volumes to correct. I am looking for a proof that is analogous to the binomial theorem, the coefficients of $$(x+y)^n$$ (each monomial comes from selecting without replacement $$m$$ 'x' vars in a monomial) and this can be viewed as combining the sub hyper-volumes of a hypercube of edge $$x+y$$.

Is there an inclusion-exclusion proof over permutations for the sub-hyper-volumes of the hyper-parallelepiped? It may well be an inductive proof where the inductive case involves $$n$$ overlapping sub-hyper-volumes.

Or more informally, here is a proof by picture using inclusion exclusion of areas for a 2x2 matrix:

Is there an analogous picture for the corresponding $$n=3$$ and the more general mapping to permutations $$n$$ arbitrary situation? I don't immediately see that adding a vector to the 2 by 2 case translates to doing three versions of 2 by 2 and adding or subtracting accordingly. How do the volumes overlap when they are sheared in three dimensions? (note that the picture proof assumes a particular ordering of magnitude of $$a, d > b, c$$; presumably the proof works with minor mods for other orderings and considering largest first).

In other words, how can the volumes correspond to terms in the determinant:

$$\begin{vmatrix} a&d&g\\ b&e&h\\ c&f&i \end{vmatrix} = a\begin{vmatrix}e&h\\f&i\end{vmatrix} - d\begin{vmatrix}b&h\\c&i\end{vmatrix} + g\begin{vmatrix}b&e\\c&f\end{vmatrix}$$

$$= aei - afh - dbi + dch + gbf - gce$$

• What exactly are you trying to prove? That the $n$-dimensional hyper-parallelipiped spanned by $n$ vectors $v_1, v_2, \ldots, v_n$ can be split into $n!$ simplices, each of which is congruent to one with vertices $0, v_1, v_2, \ldots, v_n$ ? – darij grinberg Oct 22 '18 at 14:36
• @darijgrinberg I want to prove the usual 'determinant=hypervolume' statement but each term in the sum (from one of $n!$ permutations) would correspond to some subvolume (whatever it might correspond to). They don't have to be congruent. I'd expect that they're not, but maybe you have a proof where they are. In the elementary proof by picture. the subareas are not congruent. – Mitch Oct 22 '18 at 15:32
• Ah, I see what you mean! – darij grinberg Oct 22 '18 at 15:42
• OK, slow attempt in progress... $n$ vectors have $n^n$ rectilinear 'boxes' ($2^2 = 4$, $3^3 = 27$ etc). How to get only $2! = 2$, $3! = 6$ 'boxes' to contribute to volume? wedge product is 0 for another coeff from same vector, so only permutations left..but how does this correspond to inclusion exclusion of those boxes? – Mitch Oct 23 '18 at 19:49