Lebesgue Measure and Linear Transformations II

In the first installment on the pushforward of the Lebesgue measure by a linear transformation, we restricted our attention to linear transformations defined by an orthogonal matrix and more generally, isometric affine transformations of Euclidean space. Specifically, we showed that if T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n} is an isometry, then

\lambda_{n}(T(A))=\lambda_{n}(A),\indent\forall A\in\mathcal{B}(\mathbb{R}^{n})

where \lambda_{n} denotes the n-dimensional Lebesgue measure and \mathcal{B}(\mathbb{R}^{n}) denotes the Borel \sigma-algebra on \mathbb{R}^{n}.

Today’s goal is to obtain a change-of-measure formula, when T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n} is some arbitrary affine transformation. Since the Lebesgue measure is translation-invariant, we may assume that T is linear. Proving this formula is not difficult: we only need our previous orthogonal-invariance result and some linear algebra in the form of the singular value decomposition (SVD) and the determinant function. For a review of determinants using a coordinate-free approach, check out these notes.

Using the SVD, we may write T as

\displaystyle T=UDV:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n},

where U,V:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n} are unitary operators and D:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n} is diagonalizable with respect to the standard basis. Since the determinant is a multiplicative homomorphism and unitary operators have unit determinant, we see that

\displaystyle\left|\det T\right|=\left|\det U\right|\cdot\left|\det D\right|\cdot\left|\det V\right|=\left|d_{1}\cdots d_{n}\right|,

where d_{1},\ldots,d_{n} are the diagonal entries of D. If T is non-invertible, then the image T(\mathbb{R}^{n}) is a proper subspace of \mathbb{R}^{n} and therefore of measure zero. Moreover, \det T=0, so we see that the formula holds for the singular case. Now suppose that T is invertible. We know that, for any Borel set E,


Since V(E) is again a Borel set (by continuity) and \lambda_{n}(V(E))=\lambda_{n}(E), we may assume by an approximation argument that V(E) is a rectangle R of the form


Then the measure of R under the map D is

\begin{array}{lcl}\displaystyle\lambda_{n}(D(R))=\left|(d_{1}b_{1}-d_{1}a_{1})\cdots(d_{n}b_{n}-d_{n}a_{n})\right|&=&\displaystyle\left|d_{1}\cdots d_{n}\right|\cdot\lambda_{n}(R)\\[.7 em]&=&\displaystyle\left|\det D\right|\cdot\lambda_{n}(R),\end{array}

where we use the determinant formula for a diagonal matrix to obtain the ultimate equality.

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