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edited based on Exodd's comments:

For a $m$ by $m$ matrix $A$ whose off diagonal elements are all the same $\rho$, and all the diagnoal elements are the same $\gamma + \rho$ , it can be re-written as

$$A=\gamma I+uu^{T},$$

, u is a vector of length m where each element is $\sqrt{\rho}$

Does above properties make it easier to decompose matrix $A$? e.g. Cholesky/Eigen/LU decomposition.

The only thing I can think of is eigen decomposition:

$$(A-\lambda I)\upsilon=0$$

which becomes

$$(\gamma I+uu^{T}-\lambda I)\upsilon=0$$

this leads to a new eigen problem : $$(uu^{T}-(\lambda-\gamma) I)\upsilon=0$$

if $\alpha= \upsilon - \gamma$, then $$(uu^{T}-\alpha I)\upsilon=0$$

how to go further from here ?

Can anyone kindly share some ideas?

user152503
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  • If all diagonal elements of $D$ are equal, then it is easy to obtain the eigencomposition of $A$, since it has the same eigenvectors of $uu^T$ – Exodd Dec 30 '20 at 11:02
  • @Exodd I have updated the question based on your insight. Indeed, this is a new eigen problem and the eigen vectors of $uu^{T}$ is the same as the origina A 's eigenvector – user152503 Dec 30 '20 at 11:54
  • this exact formulation has been answered countless times. You can find it and a lot of related question/answer here: https://math.stackexchange.com/questions/904926/determinant-of-a-rank-1-update-of-a-scalar-matrix-or-characteristic-polynomia – Exodd Dec 30 '20 at 12:31

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