I've asked this question [here](https://math.stackexchange.com/questions/1214449/does-y-in-mathbbrn-operatornamerankx-y-ay-2-have-zero-lebesgue) 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:


1. $n>2$
2. $x\neq0$
3. $A$ that is not a scalar multiple of the identity matrix $I\_n$
4. 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?



Lebesgue measure of set of $y\in\mathbb{R}^n$ such that $x,y,Ay$ are linearly dependent
Lemma. If $y\_1l\_2(y\_1,\dots,y\_n)-y\_2l\_1(y\_1,\dots,y\_n)\equiv 0\,$ for linear functions $l\_1,l\_2$, then $l\_i=cy\_i$ for some scalar $c$ and $i=1,2$.


Now change variables so that $x=(0,\dots,0,1)$, denote $A(y)=(l\_1(y),\dots,l\_n(y))$, $y=(y\_1,\dots,y\_n)$. If there exist $i,j$ less than $n$ such that $F(y)=y\_il\_j-y\_jl\_i$ is not identical 0, then the set you are interested in lies in a quadratic hypersurface $F=0$, thus have measure 0. If not, then by lemma there exists $c$ such that image of $A-cI\_n$ is contained in $\langle x\rangle$.


