# Meaning of signed volume

I want to understand the definition of the determinant of a $n\times n$ real matrix $A$ as the signed volume of the image of the unit cube $C'$ under the linear transformation given by $A$, i.e. $x\to Ax$. However I am failing to make sense of the words signed volume. What will be the precise definition of this? Will it be the $\int\int...\int dx_1dx_2\cdots dx_n$ over the image $C'$. I think this is an overcomplicated way to define the signed volume. Can someone suggest another way?

Thanks

• Signed volume takes into account orientation, as well as the standard notion of volume. – Alex Youcis Sep 9 '13 at 7:45
• @AlexYoucis: How do you define (a)volume, (b)orientation? – Shahab Sep 9 '13 at 8:06
• The signed volume interpretation is what motivated the definition of determinant. Since the determinant works at an algebraic level, it allows generalization to fields other than $\mathbb R$. However, the signed volume interpretation becomes inapplicable when the field is not totally ordered (in a way that is consistent with addition). – Tunococ Sep 9 '13 at 8:15
• Just in case you think the signed volume is not intuitive, pictures here might help. – Tunococ Sep 9 '13 at 8:16
• @Shahab I believe that the notion of integral follows from our general understanding of the signed volume. The value of the integral relies on chosen axes. Changing axes involves the Jacobian determinant. – Tunococ Sep 9 '13 at 8:33

This is something traditional linear algebra doesn't handle really well because it's designed to handle only vectors, scalars, and linear maps (sometimes represented by matrices) that act on vectors. The reason you're having so much trouble understanding what a "signed volume" could be is simple: signed volumes don't belong as algebraic elements in the same vector space you're used to dealing with, so it's unclear what kind of objects these are and how to describe them.

One solution to this problem is to use the geometric algebra. The geometric algebra has as part of it a wedge product of vectors that produces not another vector, but something called a 2-vector, or a bivector.

This would be written

$$C = a \wedge b$$

where $C$ is a 2-vector and $a$, $b$ are vectors.

The space of 2-vectors is in itself a vector space. 2-vectors can be added and subtracted or multiplied by scalars.

Geometric interpretation is critical here. Think of ordinary vectors ("1-vectors") as weighted directions. Each vector (except the 0 vector) has an associated unit vector (a "direction") and a unit magnitude (a "weight"). Importantly, two vectors that are scalar multiples of each other could be said to have the same direction but different weights.

Another way to think about it is to say that each unit vector represents a 1d subspace through $\mathbb R^n$.

In the geometric algebra, the vector space of 2-vectors also has unit-2-vectors. We interpret the unit-2-vectors as planes. Thus, 2-vectors are weighted planes. Each 2-vector represents an planar (or two-dimensional) subspace of $\mathbb R^n$.

A 2-vector and a 1-vector can be wedged to form a 3-vector, and this is interpreted as a "weighted" or "oriented volume". In $\mathbb R^3$, this is a vector space consisting of a single unit 3-vector and all its scalar multiples. Again, I refer to the idea that each unit $k$-vector represents a subspace. In this case, the unit 3-vector represents a subspace of $\mathbb R^3$ that is, well, $\mathbb R^3$ itself.

When we're in $\mathbb R^n$ instead, the 3-vector represents 3d subspace, but there are many such subspaces that aren't scalar multiples of each other. However, the $n$-vector has the same quality of all $n$-vectors being scalar multiples of one another, making it suitable to describe a generalized notion of volume.

Now, perhaps the real question is, what is the difference between a unit 3-vector and its additive inverse? That is, what does it mean for a subspace to be "oriented"?

A vector $v$ and its additive inverse $-v$ could be thought to point in opposite directions. This is easy to visualize, and shouldn't give you any problems. Still, I want to point out that there is some notion of orientation already present even in this case. Instead of merely saying $v$ and $-v$ are scalar multiples of each other, we can say that $v$ represents a 1d subspace one particular way, and $-v$ represents the same subspace but oriented in an opposite way.

(Any given subspace usually only admits two orientations in this manner.)

What about a 2-vector? Usually, when we talk about planes, we cheat and talk about those planes' normal vectors instead, so in some ways, people are already familiar with the idea of oriented planes. Still, it helps to imagine this intrinsically, without talking about normals. I usually imagine a sheet of paper with a counterclockwise spiral on it. The spiral defines the orientation of the sheet. Another sheet of paper that describes the same subspace could instead have spirals going clockwise.

(Clockwise and counterclockwise: again, only two orientations.)

What about a 3-vector, the "oriented volume" as it were? This is usually done using hands. You've heard of the right-hand rule, I'm sure. You can choose a 3-vector built from a set of basis vectors according to the right-hand rule, or you can choose one built from a "left-hand rule." The two represent the same subspace (all of $\mathbb R^3$), but they are nevertheless additive inverses in this system, and we have already identified such inverses for the last two cases as denoting opposite orientations.

(Right and left-hand rules. Again, only two orientations for a given subspace.)

In a general $\mathbb R^n$, there are many more such $k$-vectors, but the $n$-vector always has the quality of describing the subspace that is the whole space itself.

So far, I've shown how subspaces can be represented algebraically using the geometric algebra (as well as the exterior algebra). They can also be using matrices whose kernels are those subspace, but those matrices do not capture the important idea of orientation.

Under the geometric and exterior algebras, algebraic elements describing subspaces also carry information about those subspaces' orientations. Each subspace has only two orientations.

Now let's talk about linear algebra and the action of linear operators ("square matrices") on $k$-vectors.

There is a natural extension of a linear operator to act on these oriented subspaces. This is, in GA parlance, called "outermorphism" (after the wedge also being called an "outer" product).

Define the action of a linear map $\underline T$ on a 2-vector $C = a \wedge b)$ as

$$\underline T(C) = \underline T(a \wedge b) \equiv \underline T(a) \wedge \underline T(b)$$

That's fancy math for "$\underline T$ acts on each individual vector and then the two images are wedged". This gives a meaningful way to talk about "matrices" acting on 2-vectors.

(It's for this reason that GA users seldom talk about "matrices." The matrix expression is actually different when we talk about $\underline T$ acting on the space of 2-vectors. Having to compute new matrices for each kind of $k$-vector that might be acted upon is clumsy, and the algebra usually allows for more compact expressions of reflections, rotations, and other common operations.)

As you might expect, this extends to n-vectors also. Let $i = a_1 \wedge a_2 \wedge \ldots \wedge a_n$ be an $n$-vector, so that

$$\underline T(i) \equiv \underline T(a_1) \wedge \underline T(a_2) \wedge \ldots \wedge \underline T(a_n)$$

Remember, all n-vectors in $n$ dimensions are multiples of the unit n-vector. Or, we can just as easily say all n-vectors are multiples of $i$. So we can write

$$\underline T(i) = \alpha i$$

for some scalar $\alpha$. $\alpha$ is the determinant. All the determinant is really saying is that any $n$-vector that is put into this linear operator comes out as a scalar multiple of itself. It could be reversed in orientation. It could be scaled up or down by some factor.

The determinant is just an eigenvalue. The "eigenvector" is not a vector but a n-vector.

In summary, I have described how geometric algebra can represent subspaces in some vector space with generalized vectors called $k$-vectors. I've shown how such objects are built. I've shown that these objects have orientation associated with them. I have also described how linear operators can be made to act on these $k$-vectors as a natural, logical extension of how they act on ordinary vectors. The determinant of a linear operator is then just an eigenvalue, with "eigenvector" being the (unit) $n$-vector of the space.

• If it is of interest, I can also write a section about oriented volumes and their relevance to volume integrals. In short, when you do line integrals, you usually have a line element $d\ell$ that is a vector tangent to the curve that you integrate over, but for some accident of history, we don't usually keep track of the $n$-vector that is tangent to a manifold. One should distinguish between those tangent $n$-vectors that should appear in the integrand and the whole integral that one would use to try to find the volume of a region in the manifold. – Muphrid Oct 20 '13 at 3:10

I won't get into a lot of detail here, but there is a sort of "physical" way to at least gain some intuition on signed volumes.

Consider lifting a mass $m$ up to a height $h$. Clearly, we have imparted $mh$ units of potential energy.

Now, let us reverse the orientation of the picture, so that instead of moving from $0$ to $h$, we move from $0$ to $-h$. Clearly, in we have done $-mh$ units of work. In particular, we have done work "in the negative direction".

These quantities are the same as volumes, but one of them "adds" potential energy and the other takes potential energy away. The sign keeps track of that property.

A good intuition is the idea of continuous rigid motion, applicable to real inner product spaces.

First, we want our rigid motions to act on tuples of vectors from the space (the number of which is equal to the dimension). Second, as we fix the basepoint of these vectors at the origin, we want our rigid motions to fix the origin. Third, we want rigid motions to preserve distances. Fourth, we want a rigid motion to trace out a path in the configuration space of all ordered tuples of vectors, so our rigid motions must contain the identity map. Fifth, the rigid motion must trace out a "continuous path." Sixth, since the idea of "rotating an object" is independent of the object's actual position, and we also want our motions to be invertible; so, we want our motions to be group actions.

Thus, we can consider continuous maps $[0,1]\to{\rm O}(V)$ with $0\mapsto{\rm Id}_V$. In addition to volume of the associated parallelotopes (which are the higher-dimensional generalizations of parallelograms and parallelepipeds), an ordered tuple of vectors has more information attached to it: namely, which orbit it is in under the action of the path-connected component of the identity in ${\rm O}(V)$. The group acts transitively and there are two such components so there are two orbits. Then we can define the sign of the unit cube's orbit (associated to a chosen ordered orthonormal basis of $V$) to be $+1$ and make the sign of the other orbit $-1$. Multiply the volume and sign for the signed determinant.

Fortuitously, the signed volume can be seen to be multilinear as a function of $n$ vector arguments by examining the geometric effect of elementary matrix operations and considering degeneracy, so not only do "signed volume" and "orientation" admit an intuitive interpretation through this way, but so do does the determinant map. While the machinery is advanced for someone just learning about real vector spaces, the machinery is all very natural and once it is in place this becomes arguably "the" way to understand the idea of signed volume and orientation in this context.

In this context, signed volume is simply a term that carries slightly more information than volume alone. It's analogous to speed and velocity.

The magnitude of the determinant of a linear transformation is the number that it scales volumes in the space by. We only need to consider the unit ball however, because if you know it scales the unit ball by that number, by linearity we can simply multiply arbitrary volumes by the same number and it will tell us the volume of the scaled volume for the arbitrary figure.

However, we don't need to take the magnitude of the determinant, we can simply work with the determinant. In fact, taking the magnitude is losing some information; namely the sign. The sign actually tells us something interesting: it tells us whether the linear transformation inverts the space. To understand inversion, we can tell if a linear operator has inverted our space if we can't rotate and scale our way back. Think about the plane and the two operators:

\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -2 & 0 \\ 0 & 2 \end{bmatrix}

The first one scales any volume by 4. The second one has determinant -4, but if we take magnitudes we see that it also scales volumes by 4. The difference is that volumes have been inverted under the second operator. We can tell it's inverted because even if we re-scale by $1/2$, we cannot rotate our way back to our original coordinate system (We cannot rotate the coordinate system $(-\hat{x},\hat{y})$ to $(\hat{x},\hat{y}))$.

To put it simply, we have the following complete notion:

For a linear operator

$T : V \to V$, $A \subset V$, $B=T(A)$,

$Vol(B) = Vol(A)*|det(T)|$

If $det(T) < 0$, then $T$ inverts the space. If not, $T$ does not invert the space.

The fact that the determinant gives us this extra data of invertedness is the reason why calling it just volume would be a little unfair, so we call it signed volume.

A matrix defines a linear transformation that changes not only the volume and the shape of objects in space, but also their orientation. The magnitude of the determinant tells you by how much the volume of the object is scaled up / down. The sign of the determinant tells you whether the orientation of the object is preserved (positive sign) or changed (negative sign). A graphical illustration of these concepts can be found at https://www.statlect.com/matrix-algebra/determinant-of-a-matrix.