----- > [!proposition] Proposition. ([[scaling of Lebesgue measure under a linear transformation]]) > - Let $T \in \operatorname{End}(\mathbb{R}^{n})$ be a [[linear map|linear transformation]] of Euclidean space. > - Let $\lambda$ denote the [[Lebesgue measure]] on $\mathbb{R}^{n}$. > - Let $E$ be a [[Borel set|Borel subset]] of $\mathbb{R}^{n}$. > > Then $\lambda(TE)=|\det T| \lambda(E).$ > ^209456 > [!proof]+ Proof. ([[scaling of Lebesgue measure under a linear transformation]]) > > > [!proposition] Lemma. > > 1. [[general linear group|If]] $T \in \mathrm{GL}(\mathbb{R}^{n})$ is [[inverse linear map|invertible]], there exists $c_{T}>0$ such that $\lambda\big( TE \big)=c_{T} \lambda(E)$; > > 2. Moreover, $c_{ST}=c_{S}c_{T}$ for $S,T \in \operatorname{GL}(\mathbb{R}^{n})$. > > 3. [[general orthogonal group|If]] in fact $T \in \mathrm{O}(\mathbb{R}^{n})$, $c_{T}=1$. > > > [!proof]+ Proof of Lemma. > > Denote by $\lambda_{T}$ the [[pushforward measure|pushforward]] of $\lambda$ by $T^{-1}$, that is, $E \xmapsto{\lambda_{T}}\lambda(TE)$. Observe that since $\lambda$ is translation-invariant and [[locally finite measure|locally finite]], so is $\lambda_{T}$. Thus $\lambda_{T}=c_{T} \lambda$ for some $c_{T}>0$, by the [[uniqueness theorem for Lebesgue measure]]. We have $\begin{align} > > \lambda(ST \ E)= \lambda_{ST}(E) = \lambda_{S}(TE) = c_{S} \lambda_{T}(E) = c_{S}c_{T} \lambda(E), > > \end{align}$ > > implying $c_{ST}=c_{S}c_{T}$. Finally, if $T \in \operatorname{O}(\mathbb{R}^{n})$, then e.g. letting $E:=\{ x \in \mathbb{R}^{n}: \|x\| \leq 1 \}$ denote the closed unit ball (in particular, $TE=E$) we see $\lambda_{T}(E)=\lambda(TE)=\lambda(E)$ which implies $c_{T}=1$. > > > The claim is now that for any $T \in \operatorname{GL}(\mathbb{R}^{n})$, $c_{T}=|\det T|$. Writing $T=U SV^{\top}$ for the [[Singular Value Decomposition of an Operator|SVD]] of $T$, we have $c_{T}=c_{U}c_{S}c_{V^{\top}}=c_{S}$, where $(2)$ and $(3)$ have been applied. Write $S=\text{diag}(s_{1},\dots,s_{n})$, $s_{i}>0$. $\lambda_{S}([0,1]^{n})=\lambda\big( S([0,1]^{n}) \big)=\lambda([0,s_{1}] \times \dots \times [0,s_{n}])=\prod_{i=1}^{n}s_{i}.$ > We recognize this final value as precisely $|\det T|$. > > ^proof ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```