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)?


1 Answer 1


On your first question: Of course, $A$ has other eigenvalues. It has $n$ eigenvalues, to be specific, counting multiplicity. You have identified one. For a rank-one matrix, the other eigenvalues are all zero.

On your second question: I don't know whether it is possible in general, but it is possible if $u$ is also an eigenvector of $B$ since then the eigenvectors of $C$ are the same as those of $B$ and only the eigenvalues change. If $u$ lies in a subspace spanned by only a few of the columns of the matrix $F$ (i.e., of only a few of the eigenvectors), then the diagonalization procedure will be fairly simple since you only need to re-diagonalize those eigenvectors in this subspace, not all of them.

  • $\begingroup$ In fact, one of both $u$ and $v$ has only last non-zero entry, such as $v=(0,...,0,x)^T,\ x\neq 0$, $x$ is the given vector. If I hold this condition, what can we obtain ? $\endgroup$ Commented Oct 6, 2014 at 21:10
  • $\begingroup$ Sorry, $x$ is the given number. @Wolfgang $\endgroup$ Commented Oct 6, 2014 at 21:18
  • 2
    $\begingroup$ The eigenvector matrix $F_n$ is unchanged iff $u$ and $v$ are both eigenvectors of $B_n$ with the same eigenvalue. Indeed, if $C=F_n^*\Lambda F_n+uv^*$, then $C=F_n^*(D+\hat{u}\hat{v}^*)F_n$ for suitable $\hat{u}$ and $\hat{v}$, so we are asking essentially when $D+\hat{u}\hat{v}^*$ is diagonal. This happens iff $\hat{u}\hat{v}^*$ has a single nonzero entry on the diagonal, that is, when both $\hat{u}$ and $\hat{v}$ are multiple of the same vector of the canonical basis. $\endgroup$ Commented Oct 6, 2014 at 22:31

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