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 1



You could rewrite it as

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.