eigenvector of compositions implies eigenvector of respective functions in composition?

Question

Suppose the matrices $A, B \in Mat(n, \mathbb{F}).$ If a vector $v$ is an eigenvector of the matrix $AB,$ that is, of the composition $f_A \circ f_B,$ then is it also an eigenvector of $A$ and of $B,$ that is, of $f_A$ and of $f_B?$

I think not, for I came up with the example $$v \mapsto \lambda v + w$$ under $f_A$ for some non-zero vector $w$ that is linearly independent with respect to $v,$ and $$\lambda v + w \mapsto \lambda v$$ under $f_B.$ But, I'm not sure if my example makes any sense.

Note that $A$ and $B$ are square matrices.

Thank you!

Answer 1

I don't understand your counterexample, but the claim is false. The Euclidean basis vector $e_1$ is an eigenvector of the identity matrix. Does it follow that $e_1$ is an eigenvector of every single invertible matrix $A$ (since $AA^{-1}=I$)?

Answer 2

Not totally sure how the given example is meant to work, but consider this:

Let

$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \tag 1$

and

$K = \begin{bmatrix} 0 & 2 \\ -1 & 0 \end{bmatrix}, \tag 2$

considered as linear operators on $\Bbb R^2$, a vector space over $\Bbb R$ in the standard fashion. Then let

$L = JK = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}; \tag 3$

now $L$ clearly has eigenvectors $(1, 0)^T$ and $(0, 1)^T$ with eigenvalues $1$ and $2$, respecively, but neither $J$ nor $K$ have any real eigenvectors, since the eigenvalues of $J$ are $\pm i$, and those of $K$ are $\pm \sqrt 2 i$.