Geometrically, the determinant of a matrix is the signed volume of a unit cube after the transformation defined by the matrix is applied. However, I'm have trouble understanding what the "signed" mean here. Since effectively volumes are only positive (or zero but for now let's not worry about it). So what's the geometric meaning of a negative determinant? How should I understand the negative volume produced by applying such transformation?
[Math] What’s the geometric meaning of a negative determinant
determinantgeometrylinear algebramatrices
Related Solutions
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"?
To answer this question, I return to the case of 1-vectors, or ordinary vectors.
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.
Write $A$ as a product of the form $E_1E_2\cdots E_kDF_1F_2\cdots F_\ell$, where each $E_i$ (resp. $F_j$) is an elementary matrix corresponding to row (resp. column) switching or row (resp. column) addition, and $D$ is a diagonal matrix. You are done if you can explain why
- $\det A=\det(E_1)\cdots\det(E_k)\det(D)\det(F_1)\cdots\det(F_\ell)$ and
- the determinant of each elementary matrix is equal to the determinant of its transpose.
But these are essentially what motivate the definition of determinant: we want to define the determinant as the volume of the parallelepiped generated by the columns of a matrix. Shears do not change the volume (thus the determinant of an elementary matrix for row/column addition is $1$) and scaling in one direction changes the volume in the same proportion (thus $\det D$ is the product of $D$'s diagonal entries). Since we want to define the volume as zero in the degenerate case, it follows that together with the previous two requirements, the volume must be signed and by flipping two columns, the sign flips too (thus the determinant of an elementary matrix for row/column switching is $-1$).
Best Answer
The idea has to do with orientation, that is, which direction is right and which is left. Ultimately this has to do with which order is given for the basis elements. For example consider $\Bbb R^2$ with basis elements $(1,0)$ and $(0,1)$. We can consider $(1,0)$ to be pointing "to the right", and $(0,1)$ pointing " forward".
Putting $(1,0)$ and $(0,1)$ in a matrix gives the identity matrix whose determinant is $1$ of course.
Now switch the two basis elements. Imagine rotating the plane around the line $y=x$. Now $(1,0)$ is pointing forward but $(0,1)$ is pointing to the left.
Of course if we switch rows in the identity matrix the resulting matrix has determinant $-1$.