Determinant Scaling for General Linear Maps

From Elementary Maps to a General Linear Map

Coordinate swaps, scalings, and shears have simple effects on volume: swaps and shears preserve Lebesgue measure, while scaling one coordinate by a nonzero factor $c$ multiplies measure by $|c|$. The determinant records the combined effect of these operations. The goal here is to prove that this bookkeeping works for every invertible linear map, not just for maps given in elementary form.

Throughout, $n\geq1$, and the ambient measure space is $(\mathbb{R}^n,\mathcal{L},m)$, where $\mathcal{L}$ is the complete Lebesgue sigma-algebra and $m$ is Lebesgue measure. This measure space is sigma-finite but not finite. Sets in the result below are Lebesgue measurable, and their measures may be infinite. Matrices and vectors have real entries.

Theorem (Determinant Scaling): If $T:\mathbb{R}^n\to\mathbb{R}^n$ is an invertible linear map, then $T(E)$ is Lebesgue measurable for every $E\in\mathcal{L}$, and $$ m(T(E))=|\det T|\,m(E). $$ The equality is an equality of extended nonnegative numbers, so if $m(E)=\infty$, both sides are infinite.

We use the elementary transformation formulas established in the earlier material on coordinate permutations, diagonal scaling, and shears. A coordinate swap has measure factor $1$; scaling coordinate $i$ by $c\ne0$ has factor $|c|$; and replacing coordinate $i$ by $x_i+a x_j$, with $i\ne j$, while leaving the other coordinates fixed, has factor $1$. These formulas apply to Lebesgue-measurable sets. Each corresponding elementary matrix has determinant, in absolute value, equal to its measure factor.

Measurability Under an Invertible Linear Map

Before composing measure formulas, we must know that all intermediate images are measurable. An invertible linear map is a homeomorphism, so it sends Borel sets to Borel sets. It also sends Lebesgue-null sets to Lebesgue-null sets, as the following covering argument shows.

Let $A$ be the matrix of a linear map and let $C$ be the largest of its absolute row sums. For a cube of side length $\ell$, the image lies in a cube of side length at most $C\ell$: after subtracting the cube's center, every coordinate of an input has absolute value at most $\ell/2$, and each output coordinate has absolute value at most $C\ell/2$. Thus the image cube has volume at most $C^n\ell^n$. A null set can be covered by countably many cubes whose total volume is arbitrarily small. Their images therefore lie in cubes with total volume at most $C^n$ times that small total. This proves that the image of a null set is null; if $C=0$, the image is a singleton and is null as well.

Every Lebesgue-measurable set differs from a Borel set by a subset of a null set. The Borel part has Borel image under an invertible linear map, and the remaining part has null image by the covering argument and completeness of $\mathcal{L}$. Hence the image is Lebesgue measurable. This reasoning applies to the inverse map as well.

Reducing an Invertible Matrix to Elementary Operations

For a real $n\times n$ matrix, the elementary row operations are: swapping two rows, multiplying one row by a nonzero scalar, and adding a scalar multiple of one row to another. Each is left multiplication by an elementary matrix. The corresponding linear maps are precisely coordinate swaps, coordinate scalings, and shears.

Lemma: Every invertible real $n\times n$ matrix $A$ is a finite product of elementary matrices.

Proof. Apply Gaussian elimination to $A$. Since $A$ is invertible, its columns are linearly independent, so at each elimination stage a nonzero pivot can be found in the remaining submatrix; if necessary, swap rows to put that pivot in the pivot position. Multiply the pivot row by the reciprocal of its nonzero pivot, then use row additions to make the other entries in that pivot column zero. Repeat for the next column. After $n$ stages the resulting matrix is the identity. Every operation used is an elementary row operation, so for some elementary matrices $E_1,\ldots,E_k$ we have $E_k\cdots E_1A=I$. Each $E_j$ is invertible and its inverse is also elementary: undo the swap, use the reciprocal scalar, or subtract the row multiple. Rearranging gives $A=E_1^{-1}\cdots E_k^{-1}$, a finite product of elementary matrices. $\square$

This argument also explains why invertibility matters: a zero pivot that cannot be replaced by a nonzero pivot signals failure of full rank, so the reduction to the identity cannot be completed. The theorem here concerns only invertible maps; the singular case is separate.

Proof of the Scaling Formula

Proof. Let $A$ be the matrix of $T$. By the lemma, write $A=E_1\cdots E_k$, where each $E_j$ is an elementary matrix, and let $S_j$ be its associated linear map. Then $T=S_1\circ\cdots\circ S_k$. Starting with $E$, apply these maps from right to left. Every intermediate set is Lebesgue measurable by the measurability argument above.

For each elementary map $S_j$, its measure factor is $|\det E_j|$, by the elementary formulas stated earlier. Applying those formulas successively gives $$ m(T(E))=\left(\prod_{j=1}^k|\det E_j|\right)m(E). $$ The equality remains valid when $m(E)=\infty$, since all the factors are positive and finite. By multiplicativity of the determinant, $$ |\det A|=\left|\det(E_1\cdots E_k)\right| =\prod_{j=1}^k|\det E_j|. $$ Since $A$ represents $T$, $\det A=\det T$, and therefore $m(T(E))=|\det T|m(E)$, as claimed. $\square$

The determinant's sign records orientation, while Lebesgue measure is nonnegative. This is why the scaling factor is the absolute value of the determinant. In particular, a reflection can have determinant $-1$ but still preserves measure.

Worked Example: A Shear Followed by a Coordinate Scaling

Let $T(x,y)=(x+2y,3y)$ on $\mathbb{R}^2$, and let $E=[0,1]\times[0,2]$. Its matrix and determinant are $$ A=\begin{pmatrix}1&2\\0&3\end{pmatrix}, \qquad \det A=1\cdot3-2\cdot0=3. $$ Thus the theorem predicts $m(T(E))=3m(E)$. The rectangle has measure $m(E)=(1-0)(2-0)=2$, so the predicted image measure is $3\cdot2=6$.

For the elementary-map calculation, define $H(u,v)=(u+\tfrac23v,v)$ and $D(x,y)=(x,3y)$. Then $$ (H\circ D)(x,y)=H(x,3y)=(x+\tfrac23(3y),3y)=(x+2y,3y)=T(x,y). $$ The shear $H$ preserves measure, and $D$ multiplies measure by $|3|=3$. Hence $m(T(E))=3\cdot2=6$.

Check: The four vertices of $E$, in boundary order, map to $(0,0),(1,0),(5,6),(4,6)$. The shoelace area is $$ \frac12\left|(0\cdot0+1\cdot6+5\cdot6+4\cdot0)-(0\cdot1+0\cdot5+6\cdot4+6\cdot0)\right| =\frac12|36-24|=6. $$ This agrees with both the determinant formula and the elementary-map calculation.

Worked Example: Row Reduction in Three Dimensions

Consider the invertible matrix $$ A=\begin{pmatrix}1&2&0\\0&1&1\\0&0&2\end{pmatrix}. $$ Reduce it to the identity by the following row operations. First multiply row 3 by $1/2$, obtaining $$ A_1=\begin{pmatrix}1&2&0\\0&1&1\\0&0&1\end{pmatrix}. $$ Next replace row 2 by row 2 minus row 3: $$ A_2=\begin{pmatrix}1&2&0\\0&1&0\\0&0&1\end{pmatrix}. $$ Finally replace row 1 by row 1 minus twice row 2: $$ A_3=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I. $$

The first row operation has determinant factor $1/2$, and the next two have factor $1$. Since these operations transform $A$ to $I$, the inverse operations express $A$ as elementary maps with total measure factor $2\cdot1\cdot1=2$. Independently, $A$ is upper triangular, so $\det A=1\cdot1\cdot2=2$.

For the unit cube $Q=[0,1]^3$, $m(Q)=1$, and the theorem gives $m(A(Q))=2\cdot1=2$. The row-reduction computation gives the same factor: undoing the first operation scales the third coordinate by $2$, and undoing the other two operations uses shears, each with factor $1$. Thus the image measure is $2$.

Worked Example: A Finite Union of Rectangles

In $\mathbb{R}^2$, let $$ E=([0,1]\times[0,1])\ \cup\ ([2,3]\times[0,2]), \qquad T(x,y)=(2x,y). $$ The two rectangles are disjoint, and their measures are $1\cdot1=1$ and $1\cdot2=2$. Therefore $m(E)=1+2=3$. The matrix of $T$ is $\begin{pmatrix}2&0\\0&1\end{pmatrix}$, whose determinant is $2\cdot1-0\cdot0=2$. The theorem yields $m(T(E))=2\cdot3=6$.

Directly, the first rectangle maps to $[0,2]\times[0,1]$, of measure $2\cdot1=2$. The second maps to $[4,6]\times[0,2]$, of measure $2\cdot2=4$. These image rectangles are disjoint, so $m(T(E))=2+4=6$. This direct computation checks the theorem for a set that is not itself a single rectangle.

A Composition Pitfall

A frequent error is to compose a shear and a scaling in the wrong order or with an unchanged shear coefficient. For the map $T(x,y)=(x+2y,3y)$, let $D(x,y)=(x,3y)$. If one writes $T=H\circ D$, the shear must be $H(u,v)=(u+\tfrac23v,v)$, because $$ H(D(x,y))=H(x,3y)=(x+\tfrac23(3y),3y)=(x+2y,3y). $$ Using $H(u,v)=(u+2v,v)$ instead gives $H(D(x,y))=(x+6y,3y)$, which is a different map. For example, at $(1,1)$ the desired map gives $(3,3)$, whereas the incorrect composition gives $(7,3)$. The measure factor happens to be $3$ for both maps, so checking only the determinant or volume factor would not catch this composition error; evaluating the maps themselves does.

The determinant scaling formula is useful because a general invertible linear change of coordinates can be handled without computing the geometry of its image set directly. Once the set is Lebesgue measurable, the determinant supplies the exact factor, even when the set has infinite measure. The formula does not by itself address singular maps, for which the determinant vanishes and the image may lie in a lower-dimensional set.

Check Your Understanding

For each question, distinguish the determinant computation from the measure calculation and verify that the map being used is the intended one.

For $T(x,y)=(3x,y-4x)$, find the determinant and state the factor by which $T$ scales the measure of every Lebesgue-measurable set in $\mathbb{R}^2$.
Write the map $T(x,y)=(x+2y,3y)$ as a composition of a shear and a coordinate scaling in the order $H\circ D$. What coefficient must appear in $H$?
If $m(E)=\infty$ and $T$ is invertible, what does the determinant scaling formula say about $m(T(E))$?
For the unit cube $Q=[0,1]^3$, compute $m(A(Q))$ when $A=\begin{pmatrix}2&0&0\\0&1&1\\0&0&1\end{pmatrix}$. Identify the elementary operations represented by its diagonal scaling and shear.
Why must one establish measurability of intermediate images before applying the elementary measure formulas successively?