# Eigenvalue that depends on eigenvector

Consider the problem $$(\mathbf{1}^TAx) x=Ax,\tag{*}$$ where $$A$$ is a square matrix and $$\mathbf{1}$$ is a column vector of ones. It does not need to be of exactly this form, the main point is that we have something like an eigenvalue-eigenvector problem, where the eigenvalue $$\lambda(x)=\mathbf{1}^TAx$$ linearly depends on the eigenvector $$x$$.

This is clearly not a standard problem as the set of solutions to (*) does not form a subspace except for the case when $$x\in\mathcal{N}(A)$$. However, I believe that we still can analyse such problems from the linear algebraic perspective.

Are there any results about such problems? Existence, uniqueness of solutions? Or it should be rather treated as a general fixed-point problem?

## 1 Answer

First, find an eigenvector $$v$$ and its eigenvalue $$\lambda$$ Of course, any multiple $$cv$$ is also eigenvector with eigenvalue $$\lambda$$. What we want is to find those multiples with $$(\mathbf1^TAcv)cv=Acv,$$ i.e., $$c^2\lambda (\mathbf 1^Tv)v=c\lambda v$$ That is:

• If $$\lambda=0$$, any $$c$$ will do
• If $$\lambda\ne 0$$ and $$\mathbf 1^Tv=0$$, only the degenerate case $$c=0$$ will do
• If $$\lambda\ne 0$$ and $$\mathbf 1^Tv\ne0$$, then only $$\tag1c=\frac{1}{\mathbf 1^Tv}$$ and the degenerate case $$c=0$$ will do.

Note that $$(1)$$ boils down to normalizing our eigenvectors such that $$\mathbb 1^Tv=1$$, whereas a more "normal" normalization might perhaps attempt to make $$v^Tv=1$$.