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Let $A\in \mathbb{R}^{n\times n}$ and $v \in \mathbb{R}^n$. We recognize $Av=\lambda v$ for some scalar $\lambda$ as an eigendecomposition problem. Suppose $\mu \in \mathbb{R}^n$, and let $\odot$ denotes the Hadamard product. Does this system $Av = \lambda v+ \mu \odot v$ correspond to any generalization of eigendecomposition? Can we solve for $\lambda, v$?

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1 Answer 1

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Yes.

You could rewrite it as

$$B v = \lambda v\, $$

with $B = A - M$, $M = \textrm{diag}(\mu)$.

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