If $v$ is a column vector, then how many non-zero eigenvalues does the matrix $vv^T$ have? What are the eigenvalues? What are the corresponding eigenvectors? What are the eigenvectors corresponding to the zero eigenvalues?
[Note: This is not homework]
Let $v^T=(a_1\; a_2\;\cdots\; a_n)$ so $vv^T=(a_1v\; a_2v\;\cdots\;a_nv) $ hence the columns of $vv^T$ are lineraly dependant so $\operatorname{rank}(vv^T)=1$ (if $v\ne0$) and then $0$ is an eigenvalue with multiplicity $n-1$ and the last eigenvalue is $\operatorname{tr}(vv^T)=\sum_k a_k^2$.
Now if $(e_i)$ is the canonical basis then $e_i-e_1$, $i=2,\ldots,n$ is in the kernel of $vv^T$ so they are eigenvectors of $vv^T$ associated to the eigenvalue $0$. Finaly find vector $x=(x_1,\ldots,x_n)^T$ an eigenvector of $vv^T$ associated to the eigenvalue $\operatorname{tr}(vv^T)$ by solving $$vv^T x=\operatorname{tr}(vv^T)x$$
Since $v$ has rank $1$ then so does $vv^T$. So, there is only one non zero eigen value. Note that $vv^Tv=(v^Tv)v$ that means $v$ is the eigen vector corresponding to $v^Tv$.
