$Theorem$: If $\lambda$ is a non-zero eigen value of matrix $AB$, then same $\lambda$ is also an eigen value of matrix $BA$
$proof$:
Let $\lambda \neq 0$ be a eigen value of matrix $AB$
$\begin{align*} (AB)v &= \lambda v\\ B(AB)v&=B(\lambda v) \\ (BA)Bv &=\lambda (Bv) \end{align*}$
The point is that matrices $AB$ and $BA$ share non-zero eigen values
The dimension of $u$ and $v$ is $2$ x $1$.
The dimension of $A = uv^{T}$ is $2$ x $2$ and hence there will be two eigen values of $uv^{T}$ (repeating or non repeating)
The dimension of $v^{T}u$ is $1$ x $1$
$v^{T}u$ is dot product of $v$ and $u$ $ = 3$ and this implies eigen value of $v^{T}u$ is 3.
Since non-zero eigen values are shared by $uv^{T}$ and $v^{T}u$
So eigen values of $uv^{T}$ are $3, 0$.
Largest of the eigen value is 3 which is nothing but dot product of $u$ and $v$