Does a constant eigenvalue of a linearly parameter-dependent matrix have a constant eigenvector?

Question

Let $A(\alpha)=A_0+\alpha A_1$, with $A_0,A_1\in\mathbb R^{n\times n}$ and $\alpha\in\mathbb R$, such that there exists a $\lambda\in\mathbb C$ with the property that

for all $\alpha$: $\det(\lambda I-A(\alpha))=0$.

Does it follow that there is a constant $v\in\mathbb C^n$ which satisfies $(\lambda I-A(\alpha))v=0$ for all $\alpha$?

Answer 1

Consider $$ A(\alpha)=\begin{pmatrix}1&\alpha\\0&-1\end{pmatrix} =\begin{pmatrix}1&0\\0&-1\end{pmatrix}+\alpha\begin{pmatrix}0&1\\0&0\end{pmatrix}. $$ Clearly $\lambda=-1$ is an eigenvalue for all $\alpha$, but the corresponding eigenvector is $\binom\alpha{-2}$ up to a scalar, and therefore cannot be chosen to be independent of $\alpha$

Answer 2

suppose that $v$ be such that $(\lambda I-A(\alpha))v=0$ then we will have $\lambda v= A_0 v+\alpha A_1 v $ $ \forall \alpha \in R$ so

$ A_0 v+2 A_1 v =\lambda v$

$ A_0 v+ A_1 v =\lambda v $

and so $A_1 v=0$ and $A_0 v=\lambda v$ it will be correct if there exists a vector $v$ such that $v$ is an eigenvector of $A_0$ associated with the eigenvalue $\lambda$ and $A_1 v=0$