[Math] Scaling of Lebesgue measure under a linear transformation and the volume of a parallelepiped.

determinantlinear algebrameasure-theoryvolume

$\def\vect{\mathbf}
\def\diag{{\rm{diag}}}
\def\R{\mathbb R}
\def\vol{{\rm vol}}
\def\sign{{\rm sign}}$

This post intentionally duplicates two other threads of questions from MSE. The intention is to point out how they are connected and to give a succinct explanation.

One thread asks why the Lebesgue measure of the parallelepiped $P$ in $\R^n$ determined by vectors $v_1, \dots, v_n$ is
$$ \lambda(P) = |\det(v_1, \dots, v_n)|$$
and hence the signed volume is
$$\vol(v_1, \dots, v_n) = \det(v_1, \dots, v_n).$$

The second thread asks why Lebesgue measure scales as it does under a linear transformation, namely, given a linear transformation $T$ of $\R^n$, for all Borel sets $S$,
$$
\lambda(T(S)) = |\det(T)| \lambda(S).
$$

For the first question, see here and here.

For the second question, see here and here.

Best Answer

$\def\vect{\mathbf} \def\diag{{\rm{diag}}} \def\R{\mathbb R} \def\vol{{\rm vol}} \def\sign{{\rm sign}} $

There are two ideas involved: One is that uniqueness of Lebesgue measure as a translation invariant Borel measure implies scaling invariance under linear transformations, and in particular absolute invariance under orthogonal transformations. The other is one or another multiplicative decomposition of a linear transformation. I will use the singular value decomposition.

Theorem. Lebesgue measure $\lambda$ on $\mathbb R^n$ is the unique Borel measure that is translation invariant and locally finite (i.e. measure of compacts sets is finite), up to scaling by a positive constant.

This is contained in Rudin, Real and Complex Analysis, 3rd edition, Theorem 2.20.

Corollary

  1. If $T$ is any invertible linear transformation of $\mathbb R^n$, then there exists a positive constant $c_T$ such that for all Borel sets $E$, $\lambda(T(E)) = c_T\ \lambda(E)$.

  2. IF $S$, $T$ are invertible linear transformations $c_{ST} = c_S c_T$

  3. If $U$ is an orthogonal linear transformation, then $c_U = 1$.

Proof. For (1), note that $E \mapsto \lambda(T(E))$ is a translation invariant locally finite Borel measure. Part (2) is obvious. For part (3), it suffices to find a Borel set $S$ such that $0 < \lambda(S) < \infty$ and $U(S) = S$. But $S = \{x : ||x|| \le 1\}$ will do.

Thus Lebesgue measure is invariant under orthogonal transformations as well as under translations.

Lemma (Singular value decomposition) For any invertible matrix $A$, there exists two orthogonal matrices $W$, $V$ and a diagonal matrix $D = \diag(a_1, \dots, a_n)$, with $a_i >0$, such that $A = W D V$.

Remark: One can easily derive this from the polar decomposition and vice versa.

Corollary For any invertible $A$, $c_A = |\det(A)|$.

Proof. Write $A = W D V$, as in the Lemma, with $D = \diag(a_1, \dots, a_n)$. Then $|\det(A)| = \prod_i a_i$. On the other hand, $c_A = c_W c_D c_V = c_D$. Since $D$ applied the unit hypercube is a rectangular solid with edge lengths $a_1, \dots, a_n$, it follows that $c_A = c_D = \prod a_i = |\det(A)|$.

Lemma. The Lebesgue measure of an affine hyperplane is zero.

Proof. It suffices to consider the coordinate hyperplane perpendicular to $\vect e_n$, using translation and orthogonal invariance. Moreover, it suffices to show that the measure of any bounded subset $K$ of this coordinate hyperplane is zero. But $K$ is contained in a rectangular solid of arbitrarily small measure.

Corollary. Let $v_1, \dots, v_n$ be given and let $P$ be the parallelepiped spanned by $v_1, \dots, v_n$ . Then $\lambda(P) = |\det(v_1, \dots, v_n)|$. Moreover, the signed volume of $P$ is $\det(v_1, \dots, v_n)$

Proof. If the $v_i$ are linearly dependent then then $P$ has measure zero since $P$ is contained in a proper hyperplane, and the determinant is zero as well. Otherwise, let $A$ be the matrix $(v_1, \dots, v_n)$. Then $P$ is the the image of the unit hypercube under $A$, so $\lambda(P) = c_A = |\det(A)|$. The last statement follows from the definition of signed volume, namely $$ \vol(v_1, \dots, v_n) = \sign(\det(v_1, \dots, v_n)) \lambda(P) = \det(v_1, \dots, v_n). $$

Remark: Occasionally one sees an explanation for the scaling of Lebesgue measure or for the formula for the Lebesgue measure of a parallelepiped which invokes the change of variable formula for integration. But these explanations are circular, as the conceptual basis for the change of variable formula is the local scaling of Lebesgue measure, which depends on the global scaling of Lebesgue measure under a linear transformation.

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