Cross Products and Determinants

In summary, there is a deeper mathematical significance to the fact that the cross product is antisymmetric.
  • #1
Swapnil
459
6
Is the fact that
[tex]\vec{a}\times\vec{b} = \det \begin{bmatrix}\hat{e}_1& \hat{e}_2 & \hat{e}_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix}[/tex]

just a coincedence (ie. a mneminic device) or does it have some deep mathematical significance?

edit: Also, is the cross product, like the dot product, defined for higher dimentions?
 
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  • #2
It does indeed has a deeper mathematical significance. Note: Please take the following with a grain of salt because it very brief and I could have skipped stuff or over-simplified some things. Hopefully the gist of it is correct.

Ok, you know that whenever you do determinants you end up calculating a lot of stuff like

[tex]\det\begin{bmatrix}a & b\\c& d\end{bmatrix} = (ab - bc)[/tex]

you know, when you take the top left entry and multiply by the bottom right entry and then subtract the top right entry multiplied by the bottom left entry - continuing in this fashion more than once for larger matrices and inserting the correct plus or minus sign between calculations. You got that? (you will see later that this is like an anti-symmetric operation) For an arbitrarily large matrix you have the more generalized formula

[tex]\det M = \sum_{\pi\in S_n} \mbox{sgn}(\pi) M_{1_{\pi_1}}M_{2_{\pi_2}}\dots M_{n_{\pi_n}} = \sum_{\pi \in S_n} \epsilon_{\pi_1\dots\pi_n}M_{1_{\pi_1}}M_{2_{\pi_2}}\dots M_{n_{\pi_n}}[/tex]

which basically says you do

[tex]\det\begin{bmatrix}a & b \\ c & d \end{bmatrix} = ab - bc[/tex]

or

[tex]\det\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = aei + bfg + cdh - afh - ceg - bdi[/tex]

Now the point is that there is something called an anti-symmetric operator and it just so happens that the act of calculating the determinant of a matrix is an anti-symmetric one. Now in Exterior Algebra we use anti-symmetric operations a lot! Especially when we perform exterior products. Unlike cross products, exterior products are very, very general and work in n-dimensions. They are defined using anti-symmetric operators - which means that they involve formulas that look almost exactly like the formulas involved with determinants.

I don't know exactly what you know but if you were to gain anything from this is that the exterior product is an operator defined on vectors of arbitrary dimension, not just three. The act of performing an exterior product on two vectors involves an anti-symmetric operator (formally, it looks like this:

[tex]u \wedge v = \mathcal{A}(u \otimes v)[/tex]

where A is the anti-symmetric operator. Whenever you see the antisymmetric operator you should think of calculations like

[tex]u_i v_j - u_j v_i[/tex]

much like that in the calculation of determinants. Let's look at an example:

[tex](u \wedge v)(a,b) = \mathcal{A}(u \otimes v)(a,b)[/tex]
[tex]=\frac{1}{2}(u(a)v(b) - u(b)v(a))[/tex]

Here you can see that the wedge product of two vectors involves a calculation very similar to that of a determinant.

I guess the moral of this story is that the cross product of two 3-dimensional vectors is really an exterior product that should involve wedge products and anti-symmetrization operations, Hodge stars, etc... all the usual stuff from Exterior algebra. The reason you see it very early instead of in some advanced maths course is because the determinant is kind of like the anti-symmetrization operation, so you can replace all the hard exterior algebra stuff with easy determinants and multiplication, and all the dimensions work out nicely because the Hodge star vanishes. I guess we are just lucky that someone figured out the simplification to allow us to work with the cross product in 1st year instead of 3rd year.
Here is a brief calculation of the relation that exists between the cross product and the exterior product.

Have you heard of the Levi-Civita symbol? If so then you should agree that the cross product may be written in terms of this symbol as such:

[tex](\bold{u}\times\bold{v})_i = \epsilon_{ijk}u_j v_k[/tex]

[tex]=\frac{1}{2}\epsilon_{ijk}(u_j v_k - u_k v_j)[/tex]

[tex]=\epsilon_{ijk}(u \wedge v)_{jk}[/tex]

[tex]=\star(u \wedge v)_i[/tex]

This little calculation shows that the cross product of two vectors (in bold because they are in a 3-dimensional manifold) is related to their wedge product and the Hodge star. Note: The last equality is arises due to the fact that the Hodge star of the identity element in 3 dimensions is

[tex](\star 1)_{ijk} = \epsilon_{ijk}[/tex]
 
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  • #3
Thank you for your elaborate response. I didn't quite understand the things about Hodge Stars and anti-symmetric operators (mainly because I haven't studied Exterior Algebra yet) but I get the overall picture. :smile:
 
  • #4
Dear Oxymoron,

Thank you for offering an excellent discussion of the similarities obtaining between cross products and determinants. In my middle sixties I am finally knocking over calculus questions that I could not handle at 20. Go figure! I am, however, a little unsure as to what 'wedge products' are. When you are disposed to could you give a little gloss on that topic. Thanks again for your comprehensive and uncondescending discussion.

Ralphiep
 
  • #5


The fact that the cross product of two vectors can be represented as the determinant of a matrix is not a coincidence. It is a fundamental property of vector algebra and has deep mathematical significance. The cross product is defined as a vector that is perpendicular to both of the original vectors, with a magnitude equal to the area of the parallelogram formed by the two vectors. This geometric interpretation is closely related to the determinant of a matrix, which represents the signed volume of a parallelepiped formed by the three vectors.

Furthermore, the cross product is not limited to just three dimensions. It can be defined in any dimension, but the resulting vector will only be perpendicular to the two original vectors in three dimensions. In higher dimensions, the cross product can still be useful for calculating areas and volumes, but it may not have the same geometric interpretation as in three dimensions.

In summary, the fact that the cross product can be represented as a determinant is not just a mnemonic device, but a fundamental property with deep mathematical significance. It is also defined in higher dimensions, but its geometric interpretation may differ.
 

1. What is a cross product?

A cross product is a mathematical operation that takes two vectors in three-dimensional space and produces a third vector that is perpendicular to both of the original vectors. It is often used in physics and engineering to calculate forces and moments.

2. How is a cross product calculated?

To calculate a cross product, you need to take the components of the two vectors and use them to form a 3x3 matrix. Then, you can use the determinant of this matrix to find the magnitude of the resulting vector, and the direction can be determined using the right-hand rule. The resulting vector will be perpendicular to both of the original vectors.

3. What is the difference between a cross product and a dot product?

A cross product is a vector, while a dot product is a scalar. A cross product produces a vector that is perpendicular to the original vectors, while a dot product produces a scalar that represents the magnitude of the projection of one vector onto the other. Additionally, the cross product is only defined for three-dimensional vectors, while the dot product can be calculated for vectors of any dimension.

4. What are some applications of cross products and determinants?

Cross products and determinants have many applications in fields such as physics, engineering, and computer graphics. They are used to calculate forces and moments in rigid body mechanics, to find the direction of magnetic fields, and to determine the orientation of objects in 3D space. They are also used in computer graphics to create realistic 3D models and animations.

5. Can cross products and determinants be used with vectors in higher dimensions?

No, cross products and determinants are only defined for three-dimensional vectors. In higher dimensions, there are different mathematical operations that can be used to achieve similar results, such as wedge products and exterior derivatives.

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