Let be a matrix and let be the parallelepiped generated by the columns of . A geometrical criterion of invertibility may be:

**Property:** is invertible if and only if , where is the Lebesgue measure on .

**Proof.** Let be the columns of . Then is equivalent to for some hyperplane .

Moreover, it is known in dimension two that is the area of ; in the same way, it can be shown in dimension three that is the volume of . In fact:

**Theorem 1:** Let be a matrix. Then .

Before starting the proof, recall that Carathéodory’s extension theorem allows us to define as the unique borelian measure on satisfying for any cube. In particular, it follows the criterion:

**Lemma:** Let be a borelian measure on . If is translation invariant and finite on compact sets, then there exists such that .

**Proof of theorem 1.** If is not invertible, the theorem follows from the previous property.

For any , let be a new borelian measure on . Clearly, is translation invariant and finite on compact sets, so according to the previous lemma.

- Case 1: Suppose that is a permutation matrix.

Since , it follows that .

- Case 2: Suppose that is a diagonal matrix, say .

Since where if and otherwise,

.

- Case 3: If .

Here where is the parallelogram whose vertices are , , and , hence .

- Case 4: Consider the general case where .

According to the lemma below, can be written as a product where is either a permutation matrix or a diagonal matrix or . So, using the previous cases:

Finally, we deduce that:

.

**Lemma:** Let . Then there exist such that where is either a permutation matrix or a diagonal matrix or .

**Proof.** In order to simplify the proof, we consider only the case (writing the proof becomes painful in high dimension; however, the idea is exactly the same).

Let be an invertible matrix. We describe below an algorithm to go from to by multiplication by either a permutation matrix (operation denoted by ) or a diagonal matrix (operation denoted by ) or (operation denoted by ):

Suppose that . First, we build the first column:

And then the second column:

If then because is invertible. Applying the previous algorithm to , we get:

For an explicit example in dimension three, the previous algorithm gives:

**Remark:** We proved that for any borelian set . Therefore, tells us how evolve the volumes after applying . The sign of says wether is orientation preserving or not.