Let A \in M_n(\mathbb{R}) be a matrix and let P be the parallelepiped A \cdot [0,1]^n generated by the columns of A. A geometrical criterion of invertibility may be:

Property: A is invertible if and only if \lambda_n (P) \neq 0, where \lambda_n is the Lebesgue measure on \mathbb{R}^n.

Proof. Let v_1,\dots,v_n \in \mathbb{R}^n be the columns of A. Then \lambda_n(P) =0 is equivalent to \mathrm{span}(v_1,\dots,v_n) \subset H for some hyperplane H \subset \mathbb{R}^n. \square

Moreover, it is known in dimension two that |\det(A)| is the area of P; in the same way, it can be shown in dimension three that | \det(A)| is the volume of P. In fact:

Theorem 1: Let A \in M_n(\mathbb{R}) be a matrix. Then | \det(A) | =\lambda_n(A \cdot [0,1]^n).

Before starting the proof, recall that Carathéodory’s extension theorem allows us to define \lambda_n as the unique borelian measure on \mathbb{R}^n satisfying \lambda_n \left( \prod\limits_{i=1}^n [a_i,b_i] \right) = \prod\limits_{i=1}^n (b_i-a_i) for any cube. In particular, it follows the criterion:

Lemma: Let \mu be a borelian measure on \mathbb{R}^n. If \mu is translation invariant and finite on compact sets, then there exists \alpha \in \mathbb{R} such that \mu = \alpha \cdot \lambda_n.

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

For any A \in GL_n(\mathbb{R}), let \mu_A : B \mapsto \lambda_n (A \cdot B) be a new borelian measure on \mathbb{R}^n. Clearly, \mu_A is translation invariant and finite on compact sets, so \mu_A= \mu_A([0,1]^n) \cdot \lambda_n according to the previous lemma.

  • Case 1: Suppose that A is a permutation matrix.

Since A \cdot [0,1]^n= [0,1]^n, it follows that \mu_A([0,1]^n)=1=|\det(A)|.

  • Case 2: Suppose that A is a diagonal matrix, say A= \mathrm{diag}(a_1,\dots,a_n).

Since A \cdot [0,1]^n = \prod\limits_{i=1}^n I(a_i) where I(a_i) = [0,a_i] if a_i \geq 0 and [a_i,0] otherwise,

\displaystyle \mu_A([0,1]^n)= \prod\limits_{i =1}^n |a_i| = |\det(A)|.

  • Case 3: If A=J:= \left( \begin{matrix} \begin{matrix} 1&0 \\ 1&1 \end{matrix} & 0 \\ 0 & \begin{matrix} 1& & 0 \\ & \ddots & \\ 0 & & 1 \end{matrix} \end{matrix} \right).

Here A \cdot [0,1]^n = D \times [0,1]^{n-2} where D \subset \mathbb{R}^2 is the parallelogram whose vertices are (0,0), (1,1), (1,0) and (1,2), hence \mu_A([0,1]^n)= \lambda_2(D) \cdot \lambda_{n-2}([0,1]^{n-2}) = 1= |\det(A)|.

  • Case 4: Consider the general case where A \in GL_n(\mathbb{R}).

According to the lemma below, A can be written as a product S_1 S_2 \dots S_r where S_i is either a permutation matrix or a diagonal matrix or J. So, using the previous cases:

\begin{array}{lcl} \mu_A([0,1]^n) & = & \lambda_n(S_1 S_2 \dots S_r \cdot [0,1]^n) \\ \\ & = & \mu_{S_1} (S_2 \dots S_r \cdot [0,1]^n) \\ \\ & =& |\det(S_1)| \cdot \lambda_n(S_2 \dots S_r \cdot [0,1]^n) \\ & \vdots & \\ & = & \displaystyle \lambda_n ([0,1]^n) \cdot \prod\limits_{i=1}^r |\det(S_i) | \\ & = & |\det(A)| \end{array}

Finally, we deduce that:

\lambda_n(A \cdot [0,1]^n) = \mu_A([0,1]^n)= |\det(A)|. \hspace{1cm} \square

Lemma: Let A \in GL_n(\mathbb{R}). Then there exist S_1,\dots,S_r \in M_n(\mathbb{R}) such that A=S_1\dots S_r where S_i is either a permutation matrix or a diagonal matrix or J.

Proof. In order to simplify the proof, we consider only the case n=2 (writing the proof becomes painful in high dimension; however, the idea is exactly the same).

Let A= \left( \begin{matrix} a&b \\ c&d \end{matrix} \right) \in GL_2(\mathbb{R}) be an invertible matrix. We describe below an algorithm to go from \left( \begin{matrix} 1&0 \\ 0&1 \end{matrix} \right) to A by multiplication by either a permutation matrix (operation denoted by (1)) or a diagonal matrix (operation denoted by (2)) or J (operation denoted by (3)):

Suppose that a \neq 0. First, we build the first column:

\begin{array}{lcl} \left( \begin{matrix} 1&0 \\ 0&1 \end{matrix} \right) & \overset{(2)}{\underset{\text{if} \ c \neq 0}{\longrightarrow}} & \left( \begin{matrix} a& 0 \\ 0&c \end{matrix} \right) \overset{(3)}{\longrightarrow} \left( \begin{matrix} a & 0 \\ c & c \end{matrix} \right) \overset{(2)}{\longrightarrow} \left( \begin{matrix} a &0 \\ c & 1 \end{matrix} \right) \\ \\ & \overset{(2)}{\underset{\text{if} \ c=0}{\longrightarrow}} & \left( \begin{matrix} a&0 \\ 0 & 1 \end{matrix} \right) = \left( \begin{matrix} a & 0 \\ c & 1 \end{matrix} \right) \end{array}

And then the second column:

\begin{array}{lcl} \left( \begin{matrix} a & 0 \\ c & 1 \end{matrix} \right) & \overset{(2)}{\underset{\text{if} \ b \neq 0}{\longrightarrow}} & \left( \begin{matrix} b & 0 \\ \frac{bc}{a} & d- \frac{bc}{a} \end{matrix} \right) \overset{(1)+(3)}{\longrightarrow} \left( \begin{matrix} b & b \\ d& \frac{bc}{a} \end{matrix} \right) \overset{(2)}{\longrightarrow} \left( \begin{matrix} b & a \\ d& c \end{matrix} \right) \overset{(1)}{\longrightarrow} \left( \begin{matrix} a & b \\ c&d \end{matrix} \right) \\ \\ & \overset{(2)}{\underset{\text{if} \ b =0 }{\longrightarrow}} & \left( \begin{matrix} a & 0 \\ c & d \end{matrix} \right)= \left( \begin{matrix} a&b \\ c&d \end{matrix} \right) \end{array}

If a=0 then b \neq 0 because A is invertible. Applying the previous algorithm to \left( \begin{matrix} b & a \\ d & c \end{matrix} \right), we get:

\left( \begin{matrix} 1&0 \\ 0&1 \end{matrix} \right) \longrightarrow \left( \begin{matrix} b & a \\ d & c \end{matrix} \right) \overset{(1)}{\longrightarrow} \left( \begin{matrix} a & b \\ c & d \end{matrix} \right). \hspace{1cm} \square

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

\begin{array}{ll} \left( \begin{matrix} 1&1&0 \\ 0&1&1 \\ 1&0&1 \end{matrix} \right)= & \left( \begin{matrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 1&1&0 \\ 0&0&1 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 0&1&0 \\ 0&0&-1 \end{matrix} \right) \left( \begin{matrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 1&1&0 \\ 0&0&1 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 1&1&0 \\ 0&0&1 \end{matrix} \right)    \\ \\ & \left( \begin{matrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{matrix} \right) \left( \begin{matrix} -1&0&0 \\ 0&1&0 \\ 0&0&-2 \end{matrix} \right) \left( \begin{matrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 1&1&0 \\ 0&0&1 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 1&1&0 \\ 0&0&1 \end{matrix} \right) \left( \begin{matrix} 1&0&0 \\ 0&-1&0 \\ 0&0&1 \end{matrix} \right) \left( \begin{matrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{matrix} \right)\end{array}

Remark: We proved that \lambda_n(A \cdot B)=|\det(A)| \cdot \lambda_n(B) for any borelian set B \subset \mathbb{R}^n. Therefore, |\det(A)| tells us how evolve the volumes after applying A. The sign of \det(A) says wether A is orientation preserving or not.

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