I've asked this question here on math.stackexchange, but I have been unable to solve this yet, so I'm hoping I can get some advice here.

Consider a vector $x\in \mathbb{R}^n$ and a real $n\times n$ matrix $A$. I'm interested in the set of $y\in\mathbb{R}^n$ such that $x,y,Ay$ are linearly dependent.

To rule out trivial cases when the vectors $x,y,Ay$ are linearly dependent for any $y$, I assume that:

- $n>2$
- $x\neq0$
- $A$ that is not a scalar multiple of the identity matrix $I_n$
- the column space of $aI_n-A$ is not spanned by $x$, for any real scalar $a$.

It seems to me that the set of $y\in\mathbb{R}^n$ such that $x,y,Ay$ are linearly dependent should have zero $n$-dimensional Lebesgue measure.

Is this correct, and how would I go about proving this?

**My thinking so far:**

Let $M_x=I_n-xx'/(x'x)$ be the orthogonal projection onto the orthogonal complement of $\operatorname{span}(x)$. What I need to do, I think, is to find the measure of the set of $y\in\mathbb{R}^n$ such that $M_x y$ and $M_x Ay$ are collinear, that is, the set of $y\in\mathbb{R}^n$ such that $M_x(aI_n-A)y=0$ for some $a\in\mathbb{R}$.

Now, for any fixed $a$, the set $$S_a=\{y\in\mathbb{R}^n:M_x(aI_n-A)y=0\},$$ has zero $n$-dimensional Lebesgue measure, because $M_x(aI_n-A)\neq0$ by the assumption I've made above that the column space of $aI_n-A$ is not spanned by $x$ for any real scalar $a$.

But does the set $$\{y\in\mathbb{R}^n:M_x(aI_n-A)y=0 \text{ for some } a\in\mathbb{R} \},$$ (an uncountable union of the null spaces $S_a$ over $a$) have zero $n$-dimensional Lebesgue measure?