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If $A$ is a square $n\times n$ matrix, with $\lambda_1,\ldots,\lambda_n$ being the eigenvalues of $A$, $v_1$ being the eigenvector associated with eigenvalue $\lambda_1$, and $d$ the column vector of dimension $n$, then are the eigenvalues of $A+v_1d^T$ equal to $\lambda_1+d^Tv_1,\lambda_2,\ldots,\lambda_n$ ?

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    $\begingroup$ What do you mean by $d$ is the column vector of dimension $n$? Do you mean that $d$ is a column vector of dimension $n$? $\endgroup$ – Omnomnomnom Dec 7 '15 at 20:57
  • $\begingroup$ Yes indeed, sorry $\endgroup$ – user296769 Dec 7 '15 at 21:03
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This was my previous answer, it was wrong. See the correct statement and proof here


Hint: This is only generally true if we also know that $v_i$ is perpendicular to $d$ for $i=2,3,\dots,n$. You should be able to construct a $2 \times 2$ counterexample.

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    $\begingroup$ Game on, let me try. $\endgroup$ – user296769 Dec 7 '15 at 21:02
  • $\begingroup$ What is incorrect about this approach: A=$\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix}$ and d=$\begin{matrix} 2 \\ 0 \end{matrix}$ ? d is parallel to $v_i$ for i=2 here $\endgroup$ – user296769 Dec 7 '15 at 21:28

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