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:

  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?


1 Answer 1


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 $\langle x\rangle$.

  • $\begingroup$ Well, $F$ is defined as written: $F=y_i l_j-y_j l_i$ $\endgroup$ Commented Apr 17, 2015 at 19:36
  • $\begingroup$ 1/2 I am looking at this great answer again. It seems to me your argument can be generalized immediately to the case when $x$ is a real $n\times m$ matrix, say $X$, with assumption 1 replaced by $n>m+1$ and assumption 4 replaced by $\operatorname{col}(aI_n-A)\nsubseteq\operatorname{col}(X)$ for any real scalar $a$. The question then is whether the set of $y$ such that $\operatorname{rank}(X,y,Ay)<m+2$ has zero $n$-dim Lebesgue measure. $\endgroup$
    – mark
    Commented Jan 31, 2016 at 17:50
  • $\begingroup$ 2/2 Your argument applies if we set $X=(0_{m,n-m},I_m)^T$, because then all $(m+2)\times(m+2)$ submatrices of $(X,y,Ay)$ have zero determinant for any $y$ iff $F(y)\equiv0$ for any $i,j\leq n-m$. Can we set $X=(0_{m,n-m},I_m)^T$ without loss of generality? It seems to me we can but I'm looking for a formal argument. $\endgroup$
    – mark
    Commented Jan 31, 2016 at 17:51
  • $\begingroup$ We care only on the span of columns of $X$, in which we may choose a base consisting at most $m$ vectors, which may be settled basic units. $\endgroup$ Commented Feb 1, 2016 at 0:12
  • $\begingroup$ Thank you, very kind of you to answer my query. So, just to check that I understand correctly, we can change basis of $\mathbb{R}^{n}$ because $\mathrm{rank}(y,Wy,X)$ and the four assumptions I made are invariant to the choice of the basis of $\mathbb{R}^{n}$, right? $\endgroup$
    – mark
    Commented Feb 1, 2016 at 9:39

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