7.9  Orientation and Volume

The chapter began in §1 with a picture: the area of a parallelogram, with a sign attached, obeys a few simple rules, and those rules pinned down $ad-bc$. We have since built the determinant in every size from the same rules. It is time to return to geometry and say precisely what the determinant measures. There are two halves to the answer. Its absolute value measures volume, and its sign measures orientation, the difference between a left hand and a right hand. Both halves need care. “Volume” in ${\mathbb{R}}^7$ has no meaning until someone defines it, and “orientation” has no meaning at all over $\mathbb{C}$.

Throughout this section the field is $\mathbb{R}$.

7.9.1. What should the volume of a parallelepiped be?

For vectors ${\mathbf{a}}_1,\dots ,{\mathbf{a}}_n\in {\mathbb{R}}^n$, the parallelepiped they span is the set $$P({\mathbf{a}}_1,\dots ,{\mathbf{a}}_n):=\{t_1{\mathbf{a}}_1+\cdots +t_n{\mathbf{a}}_n:0\le t_i\le 1\text{ for every }i\}.$$ For $n=2$ it is the parallelogram with sides ${\mathbf{a}}_1,{\mathbf{a}}_2$ at the origin, and for $n=3$ a slanted box. For $n=1$ it is the segment from $0$ to ${\mathbf{a}}_1$.

We would like to say “its volume is $|det\left(\begin{array}{ccc}{\mathbf{a}}_1& \cdots & {\mathbf{a}}_n\end{array}\right)|$”. But a theorem needs two sides, and one side is missing: we have no definition of the volume of a subset of ${\mathbb{R}}^n$. School geometry supplies one for $n=2$ and $n=3$ through “base times height”, and the proper general notion, Lebesgue measure, belongs to a course in measure theory. So we proceed honestly in three steps. We define the volume of a parallelepiped by the determinant. We prove that this is the only function obeying the rules that any reasonable notion of volume must obey. And we state, without proof, that it agrees with Lebesgue measure.

The volume of a parallelepiped is the size of the determinant of its edge vectors.

In words: stack the $n$ edge vectors as the columns of a square matrix, take its determinant, and forget the sign. There must be exactly $n$ vectors in ${\mathbb{R}}^n$, so that the matrix is square.

Examples.

• The unit cube $P({\mathbf{e}}_1,\dots ,{\mathbf{e}}_n)$ has volume $|det{\mathbf{I}}_n|=1$.
• A box with edges $d_1{\mathbf{e}}_1,\dots ,d_n{\mathbf{e}}_n$ has volume $|det\mathrm{diag}(d_1,\dots ,d_n)|=|d_1\cdots d_n|$ by Theorem 7.4.6: the product of its side lengths, as it should be.
• The parallelepiped spanned by $(1,2,0)$, $(0,1,1)$, $(1,0,3)$ in ${\mathbb{R}}^3$ has volume $$|det\left(\begin{array}{ccc}1& 0& 1\\ 2& 1& 0\\ 0& 1& 3\end{array}\right)|=|1\cdot (3-0)-0+1\cdot (2-0)|=5,$$ expanding along the first row.
• Degenerate case. If ${\mathbf{a}}_1,\dots ,{\mathbf{a}}_n$ are linearly dependent, the parallelepiped is flattened into a proper subspace, and its volume is $0$ by Theorem 7.2.1. For $n=2$, $P((1,2),(2,4))$ is the segment from $(0,0)$ to $(3,6)$, with area $|4-4|=0$. This case matters: “volume zero” is the geometric face of “not invertible”.

Non-example by minimal change. Drop one vector: $(1,2,0)$ and $(0,1,1)$ span a genuine parallelogram in ${\mathbb{R}}^3$, but the matrix $\left(\begin{array}{cc}1& 0\\ 2& 1\\ 0& 1\end{array}\right)$ is $3\times 2$ and has no determinant. The clause that fails is exactly $n$ vectors in ${\mathbb{R}}^n$. The $3$-dimensional volume of that flat parallelogram would be $0$, and its area needs a different formula (Exercise (C1: Area of a triangle in space)).

Why this definition. Reordering the vectors multiplies the determinant by a sign (Theorem 7.2.1), which the absolute value removes, so the volume does not depend on the order in which the edges are listed. The absolute value is also what makes volume non-negative. The sign we threw away is not garbage: it is the orientation, the subject of the second half of this section.

7.9.2. Volume is forced by its rules

Why should anyone accept Definition 7.9.1? Because of what any notion of volume of parallelepipeds must satisfy. The pictures in §1 showed three facts about area, and they have obvious $n$-dimensional versions: stretching one edge by a factor $c$ multiplies the volume by $|c|$; sliding one edge parallel to another (a shear) does not change the volume; and the unit cube has volume $1$. The next theorem says that these three rules alone determine the volume completely. So any definition that obeys them, whether from school geometry or from measure theory, must agree with ours.

Agreement with area and volume from school. For $n=2$, let $g({\mathbf{a}}_1,{\mathbf{a}}_2)$ be the area of the parallelogram $P({\mathbf{a}}_1,{\mathbf{a}}_2)$ in the sense of school geometry, base times height. The pictures of §1 are exactly (V1)–(V3) for this $g$: stretching a side by $c$ stretches the base or the height by $|c|$; a shear ${\mathbf{a}}_1\mapsto {\mathbf{a}}_1+c{\mathbf{a}}_2$ slides the opposite side along a line parallel to the base ${\mathbf{a}}_2$, keeping base and height; and the unit square has area $1$. Hence, by Theorem 7.9.1, school area equals $|det\left(\begin{array}{cc}{\mathbf{a}}_1& {\mathbf{a}}_2\end{array}\right)|$. For $n=3$ the same argument works with “area of base times height”. What the theorem cannot supply is the geometric input itself: it takes the three rules as given.

Agreement with Lebesgue measure. In measure theory, every reasonable subset $S\subseteq {\mathbb{R}}^n$ has an $n$-dimensional volume, its Lebesgue measure. One can prove that the Lebesgue measure of $P({\mathbf{a}}_1,\dots ,{\mathbf{a}}_n)$ satisfies (V1)–(V3), and hence equals $\mathrm{vol}({\mathbf{a}}_1,\dots ,{\mathbf{a}}_n)$. We state this without proof; nothing in this book depends on it.

7.9.3. Linear maps scale volume

Applying a linear map to a parallelepiped gives another parallelepiped: by linearity, $T(\sum t_i{\mathbf{a}}_i)=\sum t_{i}T{\mathbf{a}}_i$, so $T(P({\mathbf{a}}_1,\dots ,{\mathbf{a}}_n))=P(T{\mathbf{a}}_1,\dots ,T{\mathbf{a}}_n)$. How does the volume change? Multiplicativity answers this in one line, and the answer does not depend on the parallelepiped.

This is the meaning of the determinant of a real operator: the factor by which it scales every volume. An operator with $detT=0$ squashes ${\mathbb{R}}^n$ into a lower-dimensional subspace, where $n$-dimensional volume is $0$. In multivariable calculus the same fact, applied to the linear approximation of a smooth map, becomes the Jacobian factor $|det\mathbf{J}|$ in the change-of-variables formula for integrals.

7.9.4. Orientation

Now the sign. In ${\mathbb{R}}^2$, the ordered basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$ turns counterclockwise from its first vector to its second, while $({\mathbf{e}}_2,{\mathbf{e}}_1)$ turns clockwise. No rotation carries one arrangement to the other: a rotation of the plane keeps counterclockwise turns counterclockwise. The determinant sees the difference: $det\left(\begin{array}{cc}{\mathbf{e}}_1& {\mathbf{e}}_2\end{array}\right)=1>0$ and $det\left(\begin{array}{cc}{\mathbf{e}}_2& {\mathbf{e}}_1\end{array}\right)=-1<0$.

To compare two bases of an arbitrary real space, we compare them with each other rather than with a fixed standard basis, using the change-of-coordinates matrix of Chapter 4.

Two bases have the same orientation when the matrix converting one into the other has positive determinant.

In words: write each vector of $\mathcal{B}$ in coordinates with respect to $\mathcal{C}$, put the coordinate columns side by side (Definition 4.2.1), and look at the sign of the determinant. Exactly one of the two cases occurs, because $[\mathrm{id}{]}_{\mathcal{B}}^{\mathcal{C}}$ is invertible (Theorem 4.2.1 (c)), so its determinant is non-zero (Theorem 7.5.2), and a non-zero real number is either positive or negative. The bases must be ordered: reordering a basis can change the answer.

For $V={\mathbb{R}}^n$ there is a shortcut. If $\mathbf{B}$ and $\mathbf{C}$ are the matrices whose columns are the vectors of $\mathcal{B}$ and $\mathcal{C}$, then $[\mathrm{id}{]}_{\mathcal{B}}^{\mathcal{E}}=\mathbf{B}$ and $[\mathrm{id}{]}_{\mathcal{C}}^{\mathcal{E}}=\mathbf{C}$ for the standard basis $\mathcal{E}$. By Theorem 4.2.1 (b) and (c), $[\mathrm{id}{]}_{\mathcal{B}}^{\mathcal{C}}={\mathbf{C}}^{-1}\mathbf{B}$. Since $det\mathbf{C}det{\mathbf{C}}^{-1}=det\mathbf{I}=1$ (Theorem 7.5.4), $det[\mathrm{id}{]}_{\mathcal{B}}^{\mathcal{C}}=det\mathbf{B}/det\mathbf{C}$. So $\mathcal{B}$ and $\mathcal{C}$ have the same orientation if and only if $det\mathbf{B}$ and $det\mathbf{C}$ have the same sign.

Examples.

• In ${\mathbb{R}}^2$, $({\mathbf{e}}_1,{\mathbf{e}}_2)$ and $({\mathbf{e}}_2,{\mathbf{e}}_1)$ have opposite orientations: $det{\mathbf{I}}_2=1$ and $det\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=-1$.
• In ${\mathbb{R}}^3$, $({\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3)$ and $({\mathbf{e}}_2,{\mathbf{e}}_3,{\mathbf{e}}_1)$ have the same orientation. The second matrix is the permutation matrix of a $3$-cycle, whose sign is $(-1{)}^{3-1}=1$. A cyclic shift of three vectors is two swaps.
• In $V=\mathbb{R}[x{]}_{\le 1}$, the bases $\mathcal{B}=(1,x)$ and $\mathcal{C}=(x,1+x)$ have $1=-x+(1+x)$ and $x=x+0\cdot (1+x)$, so $[\mathrm{id}{]}_{\mathcal{B}}^{\mathcal{C}}=\left(\begin{array}{cc}-1& 1\\ 1& 0\end{array}\right)$, with determinant $-1$. They have opposite orientations. Orientation makes sense in any real space, with no picture at all.
• Degenerate case, $n=1$. Bases of $\mathbb{R}$ are single non-zero numbers $(a)$, and $[\mathrm{id}{]}_{(a)}^{(b)}=(a/b)$. So $(a)$ and $(b)$ have the same orientation exactly when $a$ and $b$ have the same sign. The two orientations of a line are its two directions.

Non-example by minimal change. In ${\mathbb{R}}^2$, $(-{\mathbf{e}}_1,-{\mathbf{e}}_2)$ has the same orientation as $({\mathbf{e}}_1,{\mathbf{e}}_2)$: $det(-{\mathbf{I}}_2)=1$, and indeed it is the rotation of $({\mathbf{e}}_1,{\mathbf{e}}_2)$ by a half-turn. Change only one sign, to $({\mathbf{e}}_1,-{\mathbf{e}}_2)$: the determinant becomes $-1$, and the orientation flips. What changed is the sign of the determinant, which is the whole content of the definition. In ${\mathbb{R}}^3$, by contrast, $-{\mathbf{I}}_3$ has determinant $-1$: negating all three vectors reverses orientation.

Why this definition. Using the change-of-coordinates matrix, rather than the matrices $\mathbf{B}$ and $\mathbf{C}$ above, makes the definition work in every real space, including spaces like $\mathbb{R}[x{]}_{\le 1}$ that have no standard picture. Requiring $det>0$ rather than $det=1$ is what gives a coarse, two-valued answer: rescaling a basis vector by a positive number should not change its handedness.

The payoff is that “same orientation” really sorts all bases into two families, as the pictures suggest.

Choosing one of the two classes is called orienting $V$; its bases are then called positively oriented, and the others negatively oriented. On ${\mathbb{R}}^n$ the standard choice is the class of $({\mathbf{e}}_1,\dots ,{\mathbf{e}}_n)$. By the shortcut above, $({\mathbf{a}}_1,\dots ,{\mathbf{a}}_n)$ is positively oriented if and only if $det\left(\begin{array}{ccc}{\mathbf{a}}_1& \cdots & {\mathbf{a}}_n\end{array}\right)>0$. So the determinant of a real square matrix carries two independent pieces of information: its absolute value is a volume, and its sign is an orientation.

7.9.5. The right-hand rule

In ${\mathbb{R}}^3$, the standard orientation is the one described by the right hand: curl the fingers of the right hand from the first basis vector toward the second, and the thumb points along the third. For $({\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3)$ with the usual picture of the axes, this is the familiar arrangement below.

The cross product of $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$ in ${\mathbb{R}}^3$ is $$\mathbf{a}\times \mathbf{b}:=(a_2{b}_3-a_3{b}_2,a_3{b}_1-a_1{b}_3,a_1{b}_2-a_2{b}_1).$$ Physics defines it by the right-hand rule. Here is the linear-algebra reason the two descriptions match: $(\mathbf{a},\mathbf{b},\mathbf{a}\times \mathbf{b})$ is always positively oriented.

Expand $det\left(\begin{array}{ccc}\mathbf{a}& \mathbf{b}& \mathbf{a}\times \mathbf{b}\end{array}\right)$ along its third column (Theorem 7.6.1). The cofactors of that column are $C_{13}=a_2{b}_3-a_3{b}_2$, $C_{23}=-(a_1{b}_3-a_3{b}_1)$ and $C_{33}=a_1{b}_2-a_2{b}_1$, which are exactly the three components of $\mathbf{a}\times \mathbf{b}$. Hence $$det\left(\begin{array}{ccc}\mathbf{a}& \mathbf{b}& \mathbf{a}\times \mathbf{b}\end{array}\right)=(a_2{b}_3-a_3{b}_2{)}^2+(a_3{b}_1-a_1{b}_3{)}^2+(a_1{b}_2-a_2{b}_1{)}^2\ge 0.$$ If $\mathbf{a}$ and $\mathbf{b}$ are linearly independent, the $3\times 2$ matrix $\left(\begin{array}{cc}\mathbf{a}& \mathbf{b}\end{array}\right)$ has rank $2$, so one of its $2\times 2$ minors, which are $\pm$ the components of $\mathbf{a}\times \mathbf{b}$, is non-zero (Theorem 7.7.1). Then the determinant is strictly positive, so $(\mathbf{a},\mathbf{b},\mathbf{a}\times \mathbf{b})$ is a basis (Theorem 7.5.2) with the standard orientation. For example, ${\mathbf{e}}_1\times {\mathbf{e}}_2=(0,0,1)={\mathbf{e}}_3$, matching the picture.

7.9.6. Orientation-preserving maps

An invertible operator carries each basis to a basis. Does it keep the orientation? The answer is the same for every basis and depends only on the sign of $detT$.

An invertible operator with $detT>0$ is orientation-preserving, and one with $detT<0$ is orientation-reversing. Rotations of the plane preserve orientation and reflections reverse it (Exercise (C2: Rotations and reflections)). In ${\mathbb{R}}^3$, the map $\mathbf{x}\mapsto -\mathbf{x}$ reverses orientation, since $det(-{\mathbf{I}}_3)=-1$: a right hand, pushed through the origin, becomes a left hand.

7.9.7. Exercises

A. Check your understanding

B. Practice

C. Going deeper