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.