How do I show that for small enough $x,$ this equation $f(x) A v(x)= x v(x), f(x)\to 1, x\to 0, ||v(x)||=1, A$ fixed non-zero matrix, can't be true?

Question

I've the following:

• a fixed $n \times n$ matrix $A,$ that's a nonzero matrix and has at least one non-zero eigenvalue.
• a scalar function $f: \mathbb{R}\to \mathbb{R}, f(x)\to 1, x\to 0.$
• a vector valued function $v: \mathbb{R} \to \mathbb{R}^n, v: x\mapsto v(x) \in \mathbb{R}^n, ||v(x)||=1 \forall x\in \mathbb{R}.$
• The equality of vectors in $\mathbb{R}^n: f(x) A v(x)= x v(x) \forall x$ sufficiently small.

I'd like to show rigorously (if possible!) that for small enough $x,$ this equality won't hold. Of course if $n=1, A$ is a non-zero scalar, $v(x)= \pm 1,$ so we get $f(x)A(\pm 1) = x (\pm 1) \implies Af(x)= x$ which will have different limits ($A\ne 0$ for the left side and $0$ for the right side) as $x\to 0.$ So in this case we're done. However if $n>1, $ we can't do this. So here's what I tried so far

• Taking norms on both sides doesn't help either, as that'd yield:

$$ |f(x)||A|||||v(x)|| \ge |f(x)|||Av(x)||= |x|||v(x)|| \implies |f(x)||A|| \ge |x| \implies ||A||\ge 0, x\to 0. $$ This is useless...

• Observing that both sides have different "growth rates" w.r.t $x,$ - now I'm not very satisfied with this argument, and if possible, I'd like to rigorize it. I'm considering taking some suitable norms on both sides that'd work, but I'm running into trouble here. More particularly, I'm thinking of something like having some norm inequalities like

*$n_1(Av(x)) \ge n_2(A)n_3(v(x)), n_i's$ are suitable norms in question:

on the left and

*$|x|n_3(v(x))$

on the right, and trying to pass to $x\to 0,$ then canceling out $n_3(v(x))$ from both sides.

Note that: we can try to answer a simplified version of the question assuming $f(x)\equiv 1,$ so trying to show that:

$$A v(x)= x v(x) \forall x \in \mathbb{R}$$ won't hold?

Thank you for your help!

Answer 1

It is unclear what is given in the problem at what is the goal. Nonetheless, it is not hard to see if the equality holds and that if $f(x)\neq 0$, then $v(x)$ is an eigenvector of $A$. Being unit vector $v(x)$ has at most $2n$ possibilities. Not only that. The fraction $\frac{x}{f(x)}$ also has at most $n$ values, which are the eigenvalues of $A$. If you can clarify the set-up, we might be able to figure out what you need.