The outer product $u v^T$ more often than not appears as an operator; for instance, as Botond pointed out, if $u$ is a unit vector, then $P_u = uu^T$ is the projector onto the $u$.

There are, however, many other uses. Some of which are:

## Spectral decomposition of operators

We can express a matrix as the sum of outer products of its eigenvectors according to
$$A = \lambda_1 u_1 v_1^T + \cdots \lambda_n u_n v_n^T,$$
where $\lambda_i$ is the $i$th eigenvalue and $u_i$ and $v_i$ are the $i$th right and left eigenvectors of $A$. This is sometimes called the spectral theorem.

This is idea from linear algebra that can be extended to general spectral theory where they are used to represent rank-1 operators. Rank one operators are used a lot in quantum mechanics: for instance, one central operator in Grover's algorithm, known as *Grover's diffusion operator* is given by
$$ U_s = 2|s \rangle \langle s| - I,$$
where $|s \rangle \langle s|$ is the outer product of the vector
$|s\rangle$ in bra-ket notation.

## Rank-1 updates

In numerical algebra, we can have matrices that are updated partially. For instance, given $A^{-1}$ and two vector $u$ and $v$, we may need to compute
$$(A + uv^T)^{-1}.$$
This can be done with the Sherman-Morrison formula
which gives us a numerically cheap way to compute the inverse of a matrix when it is perturbed by a rank-1 perturbation:
$$ (A + uv^T)^{-1} = A^{-1} - \frac{A^{-1}u v^T A^{-1}}{1 + v^T A^{-1} u}.$$

A similar result holds for the determinant:
$$ \det(A +u v^T) = (1 + v^T A^{-1} u)\det(A).$$
This is called the matrix determinant lemma.

## Duality with the inner product

A personal favorite of mine is to use of the duality between the inner product and the trace of the outer product:
$$ v^Tu = \mathrm{trace}\{uv^T\}.$$

For instance, it allows for an easy (and fun) computation of the expected value of a $\chi^2$ random variable. If $X$ is $\chi^2$ distributed with $n$ degrees of freedom, then it can be seen as the sum of the squares of $n$ standard normal random variables:
$$ X = E_1 ^2 + \cdots E_n^2.$$

Let $E$ be the vector whose $i$th component is $E_i$, then $E$ is an $n$ dimensional standard normal random vector,
$$E \sim N(0,I).$$

Hence,
$$ \mathbf E [X] = \mathbf E[E^TE] = \mathrm{trace}\{\mathbf E[ E E^T]\} = \mathrm{trace}\{I\} = n.$$