Group of matrices which share a fixed eigenvector is noncommutative

Question

Let $n\in\mathbb{N}$, $n>1$, $v\in\mathbb{R}^n$. Let $X=\{A\in GL_n(\mathbb{R})\mid Av=\lambda v\text{ for some }\lambda\in \mathbb{R}\}$. Then $X$ is a noncommutative group.

It is easy to show, that $X$ is a group. But how do I show non commutativity? I can definitely construct counterexamples, but I think there should be a general argument that works for an arbitrary $v$.

Answer 1

With respect to any basis $(v,b_1,\ldots,b_{n-1})$ these are precisely the matrices of the form $$\begin{pmatrix}\lambda&u\\0&B\end{pmatrix},$$ with $u\in\Bbb{R}^{n-1}$ and $B\in\operatorname{GL}_{n-1}(\Bbb{R})$, which clearly do not all commute.

Answer 2

Since you said that you can construct counter-examples, let $A$ and $B$ be two matrices which do not commute but which have $w$ as an eigenvector. Now, let $C$ be an invertible matrix such that $Cw=v$ (we can construct such a matrix by completing $\{v,w\}$ to a basis and then writing the matrix which interchanges $v$ and $w$ but leaves all other basis vectors fixed). Consider $A'=CAC^{-1}$ and $B'=CBC^{-1}$. It is easy to check that $v$ is now and eigenvector of $A'$ and $B'$. Furthermore, if $A'B'=B'A'$ then we see that $CABC^{-1}=CBAC^{-1}$ which implies that $AB=BA$, a contradiction. Basically, we have just taken advantage of the fact that given a matrix having some eigenvector, we can change basis in a way such that another vector is an eigenvalue.

Answer 3

Matrices with distinct eigenvalues don't commute with non-diagonalisable matrices. I believe it is clear that the group contains both kinds of matrices.