# Tensor product, wedge product, Hodge product, dyad, or what?

Suppose I have two vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ in $\mathbb{R}^3$. I can regard $\mathbf{u}$ as a $3 \times 1$ matrix, and $\mathbf{v}$ as a $1 \times 3$ matrix, and then I can form their product to get a $3 \times 3$ matrix, which is sometimes denoted by $\mathbf{u} \otimes \mathbf{v}$: $$\mathbf{u} \otimes \mathbf{v} = \begin{bmatrix} u_1v_1 & u_1v_2 & u_1v_3 \\ u_2v_1 & u_2v_2 & u_2v_3 \\ u_3v_1 & u_3v_2 & u_3v_3 \\ \end{bmatrix}$$ It seems like this thing would be very useful in writing vector formulae in 3D geometry, which is what interests me.

So, my specific questions are:

(1) What is this thing called? The names "tensor product", "wedge product", and "Hodge product" all seem related somehow, but none of them seem to fit perfectly. Outer product or dyad, maybe?

(2) Where can I find a systematic account of its algebraic properties? There must be many relationships between this thing and the plain old vector dot and cross products. It's very interesting that $(\mathbf{u} \cdot \mathbf{v})\mathbf{w} = \mathbf{u} (\mathbf{v}\otimes \mathbf{w})$ for example.

(3) Is there any geometric intepretation (as there is with dot and cross products)?

I'm interested in 3D geometry, not in abstractions, so simple concrete answers would be preferable, please.

This is the tensor product of the vectors $u,v$. The tensor product operation is quite general, but in the case of $2$ vectors the simpler name "outer product" is sometimes used.

The algebraic properties you want to know come from the properties of the tensor product. For example, to see that

$$(u \cdot v)w = u(v \otimes w)$$

Pick a fourth vector $z$. Then

$$u(v \otimes w)z^{T} = (u \cdot v)(w \cdot z) = (u \cdot v)w z^{T}$$

So that both sides define the same linear map, and hence must be the same.

Lastly, regarding geometric intuition, the tensor product is a general algebraic tool that does not always have a geometric explanation. However, one can say that these objects are useful in geometry because a tensor is a linear object that behaves nicely under a change of coordinates, so it is always useful to develop geometric constructions around the framework of tensors and tensor products.

• Thanks. So, is it safe to assume that this outer product is neither a "wedge" product nor a "Hodge" product (in which case I can forget about those)? – bubba Oct 29 '13 at 0:48
• I'm still looking for geometric interpretation. I found some myself -- if $\pi$ is a plane containing the origin with unit normal $\mathbf{N}$, then, for any given vector $\mathbf{V}$, we can see that $\mathbf{V}(\mathbf{I} - \mathbf{N}\otimes \mathbf{N})$ is the the projection of $\mathbf{V}$ onto $\pi$, and $\mathbf{V}(\mathbf{I} - 2\mathbf{N}\otimes \mathbf{N})$ is the reflection of $\mathbf{V}$ through $\pi$, and so on. This is nice, because it gives us a way to express projections and reflections via matrix multiplications. – bubba Oct 29 '13 at 0:57
• I'm not really interested in proving the algebraic relationships, I'd just like to have a nice tidy list of them. But, in passing, how did you know that $u(v \otimes w)z^{T} = (u \cdot v)(w \cdot z)$? – bubba Oct 29 '13 at 1:09
• Geometry again -- there must be some relationship to distance from a point to a line, too, because these outer products show up when computing moments of inertia. – bubba Oct 29 '13 at 1:34