Is diagonal of sum matrix of rank one matrix and circulant matrix.

Question

Suppose $A=uv^T$ where $u$ and $v$ are non-zero column vectors in $\mathbb{R}^n, n\geq 3$. $\lambda=0$ is an eigenvalue of $A$ since $A$ is not of full rank. $\lambda=v^Tu$ is also an eigenvalue of $A$ since $Au=(uv^T)u=u(v^Tu)=(v^Tu)u$. Here is my questions:

(1) Are there any other eigenvalues of $A$ ?

(2) If I have another matrix $B_n$ such that $B_n = F^{*}_n\Lambda_nF_n$, and $\Lambda_n$ are diagonal matrices holding the eigenvalues of $B_n$, $F_n$ is the unitary matrix (such as the discrete Fourier-matrix). Then I want to ask:

Can sum matrix $C = B_n + A$ be diagonalized by the specified unitary matrix (In fact, $B_n$ can be the $n\times n$ circulant matrix)?

Answer 1

1. $A$ has $n-1$ zero eigenvalues, and one eigenvalue $\lambda=v^T u$ (which may also be zero), with right/left eigenvectors $u/v$.
2. The sum can be diagonalized only if $u/v$ are in the direction of eigenvectors with the same eigenvalue in $F_n$ (one if the columns).