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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)?

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  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).
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  • $\begingroup$ I am not clear to your answer 2, please give more details. In fact, one of both $u$ and $v$ have only last non-zero entry, such as $v = (0,...,0,x)$. $\endgroup$ – Hsien-Ming Ku Oct 6 '14 at 19:45
  • $\begingroup$ If that is the form of your $u$ and $v$, then if $F_n$ are Fourier matrices, then the answer to (2) is no, since any multiple of an elementary vector $e_j$ is a linear combination of all columns of a Fourier basis. $\endgroup$ – Victor Liu Oct 6 '14 at 22:29

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