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I have a real $n \times n$ symmetric matrix $\mathbf{A}_{n \times n}$ which I would like to extend with a row/column pair:

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where $\alpha$ and $\beta$ are two real numbers and $\mathbf{0}_n$ is a zero vector of size $n$.

Is there any effective algorithm which is able to compute the eigenpairs of $\mathbf{A}'$, using the eigenpairs of $\mathbf{A}_{n \times n}$?

TobiR
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  • If $\beta = 0$, it is easy. Suppose $\beta\neq 0$, since $\det(A' - \lambda I) = (\alpha - \lambda)\det(A - \lambda I - \alpha^{-1} p p^T)$, if $\alpha\neq \lambda$. Then use https://math.stackexchange.com/questions/3052997/eigenvalues-of-a-rank-one-update-of-a-matrix – Yimin Nov 29 '23 at 19:09
  • @Yimin Thanks for the tip. What is $p$ in your expression? Would this formula work when the last row and column is an arbitrary real vector? – TobiR Nov 30 '23 at 11:30
  • p is the block on your first row, second column, it is just a vector with last element as beta. Yes, it is general. – Yimin Nov 30 '23 at 15:35
  • @Yimin So $p = (0_{n-1},\beta)$? – TobiR Nov 30 '23 at 15:41
  • Yes. that is right. Just like this https://www.statlect.com/matrix-algebra/determinant-of-block-matrix – Yimin Nov 30 '23 at 15:42

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