From Determinants to Volume in \(\mathbb{R}^n\)
In \(\mathbb{R}^2\), the absolute value of a determinant gives area scaling, and in \(\mathbb{R}^3\), it gives volume scaling. The same principle holds in every finite dimension: a linear map may change the size and shape of a parallelepiped, but the absolute value of its determinant gives the factor by which its \(n\)-dimensional volume changes. The proof below uses elementary row operations and tracks their geometric effects.
Let \(n\in\mathbb{N}\), and let \(\mathbf{b}_1,\dots,\mathbf{b}_n\in\mathbb{R}^n\) be the columns of \(\mathbf{B}\in\mathbb{R}^{n\times n}\). Their parallelepiped is the set of all combinations whose coefficients lie between zero and one. Its volume is its ordinary \(n\)-dimensional Euclidean volume; if the columns do not span \(\mathbb{R}^n\), this volume is zero. The unit cube is the special case \(\mathbf{B}=\mathbf{I}_n\).
The final equality follows directly from the definition: \(T(\mathbf{B}\mathbf{t})=\mathbf{A}(\mathbf{B}\mathbf{t})=(\mathbf{A}\mathbf{B})\mathbf{t}\). Thus the scaling question is about how the volume changes when the matrix \(\mathbf{A}\) multiplies the columns of any \(\mathbf{B}\).
Geometric Effects of Row Operations
A row operation on a matrix of column vectors can be viewed as a coordinate transformation applied to each vector. In particular, left multiplication by an elementary matrix performs the same coordinate transformation on every point of a parallelepiped. The geometric volume factors are:
- Interchanging two rows interchanges two coordinates, so volume is unchanged.
- Multiplying one row by a nonzero scalar \(c\) scales one coordinate direction by \(c\), so volume is multiplied by \(|c|\).
- Adding \(c\) times one row to a different row replaces one coordinate \(x_i\) by \(x_i+cx_j\), so volume is unchanged.
For the last operation, fix all coordinates except \(x_i\). The coordinate \(x_j\) is then fixed too, so the operation translates the remaining \(x_i\)-coordinate by the fixed amount \(cx_j\). A translation does not change the length of that slice. Summing the slice lengths over the other coordinates gives the same total volume. This is the usual slicing principle for Euclidean volume. Likewise, a coordinate interchange only relabels the directions being sliced, and scaling one coordinate multiplies every corresponding slice length by \(|c|\).
Proof. A row interchange permutes two coordinates, which preserves Euclidean volume. Multiplying one row by \(c\ne0\) multiplies the length in that coordinate direction by \(|c|\), and hence multiplies volume by \(|c|\). For a row replacement \(R_i\to R_i+cR_j\), all coordinates except \(x_i\) stay fixed. On each slice obtained by fixing those other coordinates, the \(x_i\)-coordinate is translated by the fixed amount \(cx_j\); translation preserves the slice length. Slicing therefore gives the same total volume before and after the replacement. These arguments apply to parallelepipeds, including degenerate ones, whose \(n\)-dimensional volume is zero. \(\square\)
The determinant changes by corresponding factors: a row interchange changes its sign, scaling a row by \(c\) multiplies it by \(c\), and a row replacement leaves it unchanged. These are the standard determinant row-operation properties. In particular, the absolute determinant changes by the same factor \(f\) as volume. This matching is the key to the proof.
The Volume-Scaling Theorem
Proof. Apply row operations to \(\mathbf{A}\) until it is in row-echelon form \(\mathbf{R}\), and apply exactly the same operations to \(\mathbf{A}\mathbf{B}\). The result in the second case is \(\mathbf{R}\mathbf{B}\), because each row operation is left multiplication. Let \(F\) be the product of the positive volume factors for the operations. By the lemma, \(\operatorname{Vol}(P(\mathbf{R}\mathbf{B}))=F\operatorname{Vol}(P(\mathbf{A}\mathbf{B}))\). The determinant row-operation rules give \(|\det(\mathbf{R})|=F|\det(\mathbf{A})|\).
If \(\mathbf{A}\) is invertible, row reduction gives \(\mathbf{R}=\mathbf{I}_n\). Thus \(1=F|\det(\mathbf{A})|\). Also, \(P(\mathbf{I}_n)\) is the unit cube, of volume \(1\), and \(P(\mathbf{I}_n\mathbf{B})=P(\mathbf{B})\). It follows that \(\operatorname{Vol}(P(\mathbf{B}))=F\operatorname{Vol}(P(\mathbf{A}\mathbf{B}))\), so \(\operatorname{Vol}(P(\mathbf{A}\mathbf{B}))=|\det(\mathbf{A})|\operatorname{Vol}(P(\mathbf{B}))\).
If \(\mathbf{A}\) is singular, its row-echelon form \(\mathbf{R}\) has a zero row. Therefore \(\det(\mathbf{R})=0\), and the determinant row-operation rules imply \(\det(\mathbf{A})=0\). The corresponding row of \(\mathbf{R}\mathbf{B}\) is also zero, so every point of \(P(\mathbf{R}\mathbf{B})\) lies in a coordinate hyperplane and this parallelepiped has zero \(n\)-dimensional volume. Since \(F>0\), the volume relation from the lemma gives \(\operatorname{Vol}(P(\mathbf{A}\mathbf{B}))=0\). This is exactly the stated formula when \(\det(\mathbf{A})=0\). Finally, \(T(P(\mathbf{B}))=P(\mathbf{A}\mathbf{B})\), which proves the theorem. \(\square\)
The proof covers both possibilities: an invertible map preserves a full-dimensional parallelepiped’s positive volume up to a positive factor, while a singular map sends every parallelepiped into a lower-dimensional set and has zero determinant. The absolute value is necessary because volume is never negative, even though the determinant may be negative.
Worked Applications
Worked Example: Scaling a Parallelepiped in \(\mathbb{R}^2\)
Let \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), and let \(P(\mathbf{B})\) be generated by \(\mathbf{b}_1=(1,0)^{\top}\) and \(\mathbf{b}_2=(1,2)^{\top}\). Find the original and image areas.
First form the matrix of generating columns and its determinant:
In this case the area is \(2\): the parallelogram has base vector \((1,0)^{\top}\) of length \(1\) and perpendicular height \(2\). Now multiply the matrices to find the image’s generating columns:
The determinant of the map is \(\det(\mathbf{A})=2\cdot2-1\cdot1=3\). The theorem gives image area \(3\cdot2=6\). Directly, the image area is also \(\det(\mathbf{A}\mathbf{B})=2\cdot5-4\cdot1=6\), agreeing with the scaling formula. The image columns are \((2,1)^{\top}\) and \((4,5)^{\top}\), as the matrix product shows.
Worked Example: Row Reduction Tracks the Volume Factor
Find the volume of the parallelepiped that is the image of the unit cube under \(\mathbf{A}=\begin{bmatrix}0&2&0\\1&1&0\\0&0&1\end{bmatrix}\), tracking the factors used in row reduction.
Swap the first two rows, multiply the second row by \(\tfrac12\), then subtract the second row from the first:
The geometric factors for these steps are \(1\), \(\tfrac12\), and \(1\), so the total factor from the image parallelepiped to the unit cube is \(\tfrac12\). Therefore \(1=\tfrac12\operatorname{Vol}(P(\mathbf{A}))\), and the image volume is \(2\).
Check the determinant directly: expanding along the third column gives \(\det(\mathbf{A})=1\cdot(0\cdot1-2\cdot1)=-2\). Hence \(|\det(\mathbf{A})|=2\), as found by tracking the row operations.
Worked Example: A Singular Map Collapses Volume
Consider \(\mathbf{A}=\begin{bmatrix}1&2&0\\0&0&1\\0&0&0\end{bmatrix}\) acting on the unit cube in \(\mathbb{R}^3\). Its columns are \((1,0,0)^{\top}\), \((2,0,0)^{\top}\), and \((0,1,0)^{\top}\). Every image point has third coordinate zero, so the image lies in the plane \(z=0\) and has three-dimensional volume zero.
Expanding the determinant along the third row gives \(\det(\mathbf{A})=0\). The theorem predicts image volume \(|0|\cdot1=0\), consistent with the plane containment. Multiplying by each standard basis vector confirms the image columns: \(\mathbf{A}\mathbf{e}_1=(1,0,0)^{\top}\), \(\mathbf{A}\mathbf{e}_2=(2,0,0)^{\top}\), and \(\mathbf{A}\mathbf{e}_3=(0,1,0)^{\top}\).
Common Mistake: Keeping the Sign
A frequent error is to report a negative determinant as a negative volume. For example, the matrix in the second worked example has determinant \(-2\). The incorrect conclusion “the image volume is \(-2\)” confuses signed determinant with ordinary volume. The correction is to take the absolute value: \(|-2|=2\). The sign records orientation information; the \(n\)-dimensional volume is nonnegative.
Another useful distinction is what is being scaled. The theorem concerns the volume of a parallelepiped generated by \(n\) vectors, not just the unit cube. If the starting parallelepiped has volume \(V\), its image has volume \(|\det(\mathbf{A})|V\). The matrix \(\mathbf{B}\) records the starting shape and size; \(\mathbf{A}\) contributes the scale factor.
Why the Formula Is Useful
The volume formula gives a geometric interpretation of the determinant in any dimension and clarifies what a zero determinant means: the map collapses every \(n\)-dimensional parallelepiped into a set of zero \(n\)-dimensional volume. It also separates two pieces of information. The magnitude \(|\det(\mathbf{A})|\) is the volume factor, while the sign of \(\det(\mathbf{A})\) distinguishes orientation. When solving problems, first identify the original volume and then multiply by the absolute determinant; this avoids confusing the size of a particular parallelepiped with the scale factor of the map.
Check Your Understanding
Use the theorem and the row-operation volume factors to check your reasoning.
- If \(\det(\mathbf{A})=-5\) and a parallelepiped has volume \(3\), what is the volume of its image?
- What volume factor corresponds to multiplying one row by \(-4\)? Why is it not negative?
- Explain why a singular \(3\times3\) matrix maps the unit cube to a set of zero three-dimensional volume.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\2&3\end{bmatrix}\), find the area scale factor and state whether the determinant sign affects area.
- In the theorem, what role does \(\mathbf{B}\) play, and which matrix determines the volume scale factor?
