On Wikipedia page on vector multiplication, they gave several ways to "multiply" two vectors, the most well known being the dot product and the cross product

I am curious whether we can multiply two vectors as is done with polynomials.

For example, let $u = a_1 \hat i + b_1 \hat j + c_1 \hat k$ and $v = a_2 \hat i + b_2 \hat j + c_2 \hat k$

Then a possible product between u and v would be $$u \times v = a_1a_2 \hat i \hat i + a_1 b_2 \hat i \hat j + a_1 c_2 \hat i \hat k + ...$$ you get the idea

Is this operation well defined? Why is this operation seldom used?

  • $\begingroup$ But that's exactly the cross product. $\endgroup$ – Aldon May 6 '15 at 21:39
  • 2
    $\begingroup$ No, this is not well defined. The central problem is that you have not defined what $\hat{i}\hat{j}$ means. $\endgroup$ – Emily May 6 '15 at 21:39

Summary: Not only was the first product of vectors defined exactly this way, but the dot and cross products are just parts of this first product -- the product of quaternions AKA the Hamilton product.

You might be interested to know that $i$, $j$, and $k$ were introduced by William Hamilton as the imaginary parts of quaternions. And he developed the rules for multiplying his numbers with exactly that distributive property (and some other nice properties, like associativity) in mind.

For some background, recall that complex numbers are written in Cartesian form as $a+bi$, where $a, b$ are real numbers and $i$ is the imaginary unit such that $i^2=-1$.

Well, complex numbers are extremely useful in physics, but Hamilton wanted to go one step further -- he wanted to generalize complex numbers to $3$ dimensions. Then we could use these generalized complex numbers to represent points in $3$-D space.

He never found a $3$-D generalization of complex numbers but he did find a $4$-D generalization -- the quaternions.

Quaternions are numbers of the form $a+bi+cj+dk$ where $a, b, c, d$ are real numbers, and $i, j, k$ are distinct imaginary units. These imaginary units have definitions similar to the imaginary unit in the complex numbers: $i^2=j^2=k^2=-1$. But you also need some way of figuring out what things like $ij$ means. To this end, Hamilton figured out you only need to add one more rule $ijk=-1$. Then you can figure out what $ij$ is by multiplying both sides of $ijk=-1$ by $k$. Then you get $$(ijk)k = (-1)k \\ ij(k^2)=-k \\ ij(-1) = -k \\ ij = k$$

Hamilton is also the guy who introduced the words "vector" and "scalar". To Hamilton scalars were the real parts of his quaternions and vectors were the imaginary parts. For instance, if you had $z= 4 + 5i + 6j -7k$, then the scalar part of $z$ is $4$ and the vector part is $5i+6j-7k$.

Quaternions are not only the progenitors of modern vectors, but also of the dot and cross products. Let's look at the Hamilton product of two quaternions with 0 scalar parts:

$$(a_1i + a_2j + a_3k)(b_1i+b_2j + b_3k) \\ = a_1b_1(ii) + a_1b_2(ij) + a_1b_3(ik) + a_2b_1(ji) + a_2b_2(jj) + a_2b_3(jk) + a_3b_1(ki) + a_3b_2(kj) + a_3b_3(kk) \\ = -a_1b_1 + a_1b_2(k) + a_1b_3(-j) + a_2b_1(-k) - a_2b_2 + a_2b_3(i) + a_3b_1(j) + a_3b_2(-i) - a_3b_3 \\ = -(a_1b_1 + a_2b_2 + a_3b_3) + \left[(a_2b_3-a_3b_2)i + (a_3b_1-a_1b_3)j + (a_1b_2 - a_2b_1)k\right]$$

But you should recognize this part part: it's just $-\mathbf a \cdot \mathbf b + \mathbf a \times \mathbf b$. After some study it was discovered that these two subproducts of the Hamilton product are very, very useful.

You may be wondering why we don't use quaternions nearly as much as vectors today -- it's mostly due to the work of some very influential mathematicians and physicists like Oliver Heaviside and Josiah Gibbs.

In particular, Josiah Gibbs really did not like Hamilton's approach to quaternions. So he spent several years developing a calculus on just the vector parts of quaternions -- which he saw as much more useful than the full quaternion analysis.

Gibbs produced a short treatise on his methods but was unable (or unwilling) to write up a full textbook on the subject. A new graduate student from Harvard, Edwin Wilson, was convinced by his advisor that he should write a textbook based on Gibbs' treatise. It was entitled Vector Analysis and it was one of the most important textbooks in the history of mathematics. Not because it was so visionary per se, but because the methods, notations, and ideas of this book have become so fundamental to modern mathematics that we teach it to high schoolers (and of course, every university student was has to take vector calculus, linear algebra, or classical mechanics -- so pretty much all of them).

  • $\begingroup$ Beautiful answer! Two quick questions: 1. Why doesn't three dimensional complex number exist? one could argue that the quaternion without real part is a three dimensional complex number. 2. Could you cite a source for Hamilton's introduction of the word "scalar" and "vector" that would be wonderful. $\endgroup$ – Carlos - the Mongoose - Danger May 10 '15 at 3:30
  • $\begingroup$ @IllegalImmigrant $(1)$ Hurwitz's theorem tells us that all normed division algebras over $\Bbb R$ are isomorphic to either the real numbers ($1$D), the complex numbers ($2$D), the quaternions ($4$D), or the octonions ($8$D). You can construct a Cayley-Dickinson table for $3$D numbers but they'd be trivial extensions of the complex numbers -- not really useful in their own right. Here, ... $\endgroup$ – user137731 May 10 '15 at 14:35
  • $\begingroup$ here, and here are relevant questions from this site that you may want to look at. $\endgroup$ – user137731 May 10 '15 at 14:37
  • $\begingroup$ $(2)$ A set the of quaternions without real parts is not closed under multiplication (just look at my answer to see that a scalar part shows up). On the other hand, people like Gibbs decided it would be better to just get rid of the Hamilton product and treat the imaginary parts of quaternions as objects in their own right. Thus modern vector algebra. $\endgroup$ – user137731 May 10 '15 at 14:38
  • $\begingroup$ $(3)$ Earliest Known Uses of Some of the Words of Mathematics -- just scroll down to "Vector and Scalar". $\endgroup$ – user137731 May 15 '15 at 23:31

All products basically work this way, with just different ideas or interpretations about what quantities like $\hat i \hat i$ should mean.

For instance, the dot product sets $\hat i \hat j = 0$, and similarly for any other pair of different standard basis vectors. Otherwise, $\hat i \hat i = \hat j \hat j = \hat k \hat k = +1$.

The cross product sets $\hat i \hat i = 0$ and $\hat i \hat j = \hat k$, and so on.

The tensor product (denoted $\otimes$) interprets these as forming a linear map: $\hat i \otimes \hat j$ becomes a function of two vectors $f, g$ such that $(\hat i \otimes \hat j)(f,g) = (\hat i \cdot f)(\hat j \cdot g)$.

Exterior algebra's wedge product (denoted $\wedge$) sets $\hat i \hat i = \hat j \hat j = \hat k \hat k = 0$ and $\hat i \hat j = -\hat j \hat i$ and similarly, along with associativity. This has properties like the cross product, but can be more convenient and useful for higher dimensional spaces.

The "geometric product" from clifford algebra sets $\hat i \hat i = \hat j \hat j = \hat k \hat k = +1$ instead, but is otherwise like the wedge product. This unifies the dot product behavior with the wedge product.

We use different symbols for these products to avoid getting them confused, but they all generally follow the distributive property and work the way you want them to work. They just impose different multiplication laws for certain special products, reducing the operations to what you already know.


I think you are describing the tensor product: http://en.wikipedia.org/wiki/Tensor_product


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