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Let $\bf A$ be a $n \times n$ symmetric positive semidefinite matrix whose first column is denoted by ${\bf a}_1$. We define a new matrix,

$$ {\bf B} = \begin{bmatrix} a_{11} & {\bf a}_1^T \\ {\bf a}_1 & {\bf A} \end{bmatrix} $$

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

Let $\lambda_k$ and $\mu_k$ denote 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?

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  • $\begingroup$ 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 $n \rightarrow \infty$. $\endgroup$ Commented Jul 24, 2014 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 $0 < \rho < 1$. $\endgroup$ Commented Jul 24, 2014 at 19:05
  • $\begingroup$ To sum up this discussion, there's absolutely no natural scale available here without further assumptions. $\endgroup$ Commented Jul 24, 2014 at 19:08
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    $\begingroup$ The eigenvalues of B interlace those of A, see for example members.upc.nl/w.haemers/interlacing.pdf $\endgroup$ Commented Jul 25, 2014 at 0:10

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

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