As I understand it, the power iteration is a method used to approximate the eigenvalue with the largest magnitude. It has a convergence rate $\frac{\lambda_{\max}}{\lambda_{\min}}$.
I am given the following question:
Given that we have a matrix $n \times n$ with rank $1$, this implies that it has a single eigenvalue given by $u v^T$. If we want to apply the power iteration to that matrix, what eigenvector will it converge to and how many iterations are required for it to converge exactly to that eigenvector?
My attempt
Given that $A$ has rank $1$ and it has an eigenvalue $u v^T$ $\Rightarrow A$ has eigenvalues $= 0$ with multiplicity $(n-1)$ following $\mbox{rank}(A) = 1 \Rightarrow \mbox{null}(A) = n-1$ Then the power iteration will find eigenvector $u$ but what about its convergence rate? Is it infinity?
There is probably a typo. The matrix $A$ is $uv^\top$, but its nonzero eigenvalue is $v^\top u$, corresponding to the eigenvector $u$.
Note that the column space of $A$ is one-dimensional and spanned by $u$. So the power method finds the eigenvector in one step.