From Edge Vectors to Volume
A parallelepiped in \(\mathbb{R}^n\) is determined by a base point and \(n\) edge vectors. Its volume depends on how those vectors stretch and tilt the coordinate unit cube together, not on the product of their lengths alone. The determinant records this combined effect: its absolute value is the volume factor, while its sign records orientation.
Throughout, volume means Lebesgue measure \(m\) on \((\mathbb{R}^n,\mathcal{L},m)\). This measure space is not finite, but the parallelepipeds considered here are bounded and have finite measure. Given \(a\in\mathbb{R}^n\) and edge vectors \(v_1,\ldots,v_n\in\mathbb{R}^n\), define the closed parallelepiped
Let \(A\) be the \(n\times n\) matrix whose \(j\)-th column is \(v_j\). The map \(t\mapsto a+At\) is continuous, and \([0,1]^n\) is compact, so \(P\) is compact and hence Lebesgue measurable. The base point translates the set but does not affect its volume.
The justification uses the determinant-scaling result for Lebesgue measure established in the earlier material on general linear maps: if \(T:\mathbb{R}^n\to\mathbb{R}^n\) is linear and \(E\in\mathcal{L}\), then \(T(E)\) is Lebesgue measurable and \(m(T(E))=|\det T|m(E)\). Translation invariance also holds. The unit cube has measure \(m([0,1]^n)=1\).
Proof. If \(\det A\ne0\), then \(P=a+A([0,1]^n)\). By determinant scaling and translation invariance, \(m(P)=|\det A|m([0,1]^n)=|\det A|\). If \(\det A=0\), the linear map \(A\) is singular. By the result on images under singular linear maps established earlier, \(A([0,1]^n)\) has Lebesgue measure zero. Translation invariance gives \(m(P)=0=|\det A|\). These cases cover every matrix \(A\). \(\square\)
The formula uses Lebesgue measure, so volume is always nonnegative. A negative determinant does not mean that the set has negative volume; it means that the ordered edge vectors give the opposite orientation. A zero determinant means that the edge vectors are linearly dependent, so the image has no \(n\)-dimensional volume.
Worked Determinant Calculations
Worked Example: A Parallelogram in the Plane
In \((\mathbb{R}^2,\mathcal{L},m)\), take \(a=(0,0)\), \(v_1=(3,1)\), and \(v_2=(1,2)\). Using the edge vectors as columns gives
Therefore the parallelogram has Lebesgue measure \(m(P)=|5|=5\). The determinant is positive, so this order of the two edges gives positive orientation.
Check by base and height. The base length is \(\|v_1\|=\sqrt{3^2+1^2}=\sqrt{10}\). The dot product is \(v_1\cdot v_2=3\cdot1+1\cdot2=5\), and \(\|v_2\|^2=1^2+2^2=5\). Thus the squared length of the component of \(v_2\) perpendicular to \(v_1\) is \(5-5^2/10=5-5/2=5/2\), so the height is \(\sqrt{5/2}\). The base-times-height area is \(\sqrt{10}\sqrt{5/2}=\sqrt{25}=5\), agreeing with the determinant calculation.
Worked Example: A Three-Dimensional Parallelepiped
In \((\mathbb{R}^3,\mathcal{L},m)\), let \(a=(0,0,0)\) and choose \(v_1=(1,2,0)\), \(v_2=(0,1,1)\), and \(v_3=(2,0,1)\). With the vectors as columns,
Hence \(m(P)=5\). The determinant is positive, so the listed edge order has positive orientation.
Check by a scalar triple product. The cross product of the second and third edges is \(v_2\times v_3=(1\cdot1-1\cdot0,\ 1\cdot2-0\cdot1,\ 0\cdot0-1\cdot2)=(1,2,-2)\). Taking its dot product with the first edge gives \(v_1\cdot(v_2\times v_3)=1\cdot1+2\cdot2+0\cdot(-2)=5\). Its absolute value is \(5\), as required.
Worked Example: Negative Orientation and Coordinate Order
In \((\mathbb{R}^3,\mathcal{L},m)\), take the edge vectors \(v_1=(1,0,1)\), \(v_2=(0,2,1)\), and \(v_3=(1,1,0)\). Their column matrix and determinant are
The parallelepiped volume is \(|-3|=3\), not \(-3\). The negative sign records the orientation of the ordered edge vectors.
Now reorder the coordinates from \((x,y,z)\) to \((z,y,x)\). This swaps the first and third rows of the matrix, giving
The determinant sign has changed, but the absolute value remains \(3\): permuting coordinate axes preserves volume. Likewise, swapping two edge vectors swaps two columns, changes the determinant sign, and leaves the geometric parallelepiped unchanged. For example, swapping \(v_1\) and \(v_2\) only relabels the parameters \(t_1,t_2\), which range over the same interval \([0,1]\). Thus the volume is unchanged.
Check: the coordinate swap is a reflection with determinant \(-1\), so determinant scaling multiplies volume by \(|-1|=1\). The computed volumes before and after the swap are both \(3\).
Degeneracy and Consistent Coordinates
The determinant also detects whether the parallelepiped collapses into a lower-dimensional set. In \(\mathbb{R}^n\), a square matrix has determinant zero exactly when its columns are linearly dependent. In that case its image is contained in a proper linear subspace, and the singular-map result gives zero \(n\)-dimensional Lebesgue measure. This does not say that the image is empty: it can be a segment, a planar region in \(\mathbb{R}^3\), or another nonempty lower-dimensional set.
Worked Example: Dependent Edges Give Zero Volume
In \((\mathbb{R}^3,\mathcal{L},m)\), use \(v_1=(1,2,0)\), \(v_2=(2,4,0)\), and \(v_3=(0,1,1)\), with \(a=(0,0,0)\). Since \(v_2=2v_1\), the edge vectors are dependent. The matrix calculation confirms this:
More explicitly, a point of the parallelepiped is \((t_1+2t_2,\ 2t_1+4t_2+t_3,\ t_3)\). Writing its coordinates as \((x,y,z)\), we have \(y=2x+z\). Thus the whole set lies in the plane \(y=2x+z\), and the singular-map result gives \(m(P)=0\).
Check: the second column is exactly twice the first, so the columns cannot span \(\mathbb{R}^3\); the determinant and the volume formula both give zero.
For calculations, fix an order for the coordinate components and place each edge vector consistently as a column. If instead each edge is written as a row, the matrix is the transpose of the column matrix and has the same determinant. Either convention works when applied consistently. The potential error is to mix conventions—for example, to put some edge vectors in rows and others in columns—or to change coordinate order in only part of the calculation. A coordinate permutation changes the determinant by the sign of that permutation, but its absolute value, and therefore volume, is unchanged.
A Common Mistake: Keeping the Sign as Volume
For the third worked example, a wrong conclusion would be: “The determinant is \(-3\), so the parallelepiped has volume \(-3\).” This confuses signed orientation with Lebesgue measure. The correction is to apply the absolute value in the volume formula: \(m(P)=|\det A|=|-3|=3\). A second common error is to multiply edge lengths as if the edges were perpendicular. In the planar example, that gives \(\sqrt{10}\sqrt{5}=\sqrt{50}\), not the correct area \(5\); the product of lengths is area only when the two edge vectors are perpendicular.
The determinant calculation is useful precisely because it handles tilted, reflected, and degenerate configurations in one formula. For a full-dimensional parallelepiped, its absolute value gives the exact volume regardless of translation or orientation. For dependent edge vectors, it correctly returns zero. When checking a calculation, verify the matrix convention, calculate the signed determinant first, and take the absolute value only when reporting volume.
Check Your Understanding
Use the edge vectors as columns unless a question specifies another consistent convention.
- In \(\mathbb{R}^2\), find the volume of the parallelogram generated by \((2,1)\) and \((1,3)\). What is the orientation in this order?
- Explain why swapping two edge vectors changes the sign of the determinant but not the parallelepiped’s volume.
- In \(\mathbb{R}^3\), what does a zero determinant imply about the edge vectors and the three-dimensional Lebesgue measure?
- If the order of two coordinate axes is swapped, what happens to the determinant and to the volume?
- Give a condition under which multiplying the lengths of the edges gives the volume, and explain why it does not work for arbitrary edge vectors.