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 $\lambda, v$?
1 Answer
Yes.
You could rewrite it as
$$B v = \lambda v\, $$
with $B = A - M$, $M = \textrm{diag}(\mu)$.