# Meaning of juxtaposition of vectors

I came across some notation that I can't quite understand:

$$\hat{r}\hat{r} - \textbf{1}_3$$

where $\textbf{1}_3$ is the 3$\times$3 identity matrix, $\hat{r}$ is a unit 1$\times$3 vector, and the result is supposed to be a 3$\times$3 matrix. What is the operation (implied) in the juxtaposition $\hat{r}\hat{r}$?

The source of this notation is equations (6) and (10) in this paper.

-

## 2 Answers

The operation is the tensor product, an operation producing many components (products of every component from the left vector and every component from the right vector is remembered). In this tensor notation, the inner product (which involves the summation of 3 products) would have to be explicitly indicated by a dot, $a \cdot b$.

The whole expression ${\bf \hat r\hat r - 1}$ is supposed to be a matrix, i.e. object with two vector-like indices $M_{ij}$. In terms of components, $$M_{ij} = \hat r_{i} \hat r_{j} - \delta_{ij}$$ where the $\delta$ term is the Kronecker delta – a representation of the unit matrix that is equal to $1$ for $i=j$ and $0$ otherwise. Also, $\hat r_i \equiv r_i / r$.

If one looks what $M_{ij}$ means e.g. for $r=(0,0,z)$ for $z\gt 0$, in the positive $z$-direction, he finds out that $M_{ij}={\rm diag}(-1,-1,0)$.

Just to be sure, the indices $j,k$ in the equation 6 of the paper the OP mentioned aren't vector indices. They're labels identifying different objects/sources at different locations of space.

-
Great answer, thanks! And yes - I got that the indices refer to different objects rather than matrix components. Did you happen to catch the intuition behind equation (6)? It supposed to represent how one dipole (j) 'talks' to another dipole (k) - but the expression looks complicated and there's no source or explanation for it. –  alexvas Dec 26 '12 at 9:34
Hi, yup. The $rr-1$ "complicated" structure of the terms in the dipoles comes from the differentiation (gradient) of the electric field of an electric charge/monopole which is much simpler. The equation 6 over there also has some plane waves etc. See en.wikipedia.org/wiki/Electric_dipole for some related but simpler formulae concerning dipoles... –  Luboš Motl Dec 26 '12 at 9:40

I would guess it's multiplying the vector with its tranpose, i.e. $\hat r (\hat r ^\intercal)$

so if $\hat r = \begin{pmatrix}1\\0\\0\end{pmatrix}$, then $\hat r \hat r = \begin{pmatrix}1\\0\\0\end{pmatrix} \begin{pmatrix}1 & 0 & 0\end{pmatrix} = \begin{pmatrix}1 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{pmatrix}$

-
This was my original thought but I wanted to be sure. –  alexvas Dec 26 '12 at 9:48