Let ${\bf A}$ be a size $n \times n$ symmetric positive semidefinite matrix with the first column being ${\bf a}_1$. If we define a new matrix, \begin{align} {\bf B} = \left[\begin{array}{cc} a_{11} & {\bf a}_1^T \\ {\bf a}_1 & {\bf A} \end{array}\right] \end{align}

How are the eigenvalues of ${\bf B}$ related to those of ${\bf A}$?

Let $\lambda_k$ and $\mu_k$ being the eigenvalues of ${\bf A}$ and ${\bf B}$, respectively. My simulations suggest that $\frac{\lambda_k}{n}$ and $\frac{\mu_k}{n+1}$ are very similar as $n \to \infty$. Can we prove this?

  • $\begingroup$ Can you be more specific about how similar you'd like them to be? If $A$ has an $n$ in the top-left corner and zeroes elsewhere, then $\lambda_{max}(A)/n = 1$, but $\lambda_{max}(B)/(n+1) = 2n/(n+1) \rightarrow 2$ as $n \rightarrow \infty$. $\endgroup$ – Nathaniel Johnston Jul 24 '14 at 18:58
  • $\begingroup$ You are right. Probably the observation that $\lambda_k/n$ converges to $\mu_k/(n+1)$ is only true for some special cases. The example I tried was $a_{ij} = \rho^{|i-j|}$, where $0 < \rho < 1$. $\endgroup$ – Jingxian Wu Jul 24 '14 at 19:05
  • $\begingroup$ To sum up this discussion, there's absolutely no natural scale available here without further assumptions. $\endgroup$ – Christian Remling Jul 24 '14 at 19:08
  • 3
    $\begingroup$ The eigenvalues of B interlace those of A, see for example members.upc.nl/w.haemers/interlacing.pdf $\endgroup$ – Brendan McKay Jul 25 '14 at 0:10

Since this is a rank-one modification, you can apply the result of 10.1007/BF01396012.


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