Showing that a non-hyperbolic linear map is not structurally stable

Let $A \in GL(\mathbb{R}^n)$ such that $A$ is not hyperbolic (i.e $A$ has at least an eigenvalue $\lambda$ with $| \lambda | = 1$ ). Show that
$$x \mapsto Ax$$ is not structurally stable.

A $C^r$ map $f$ is $C^m$ -structurally stable ($1 \leq m \leq r$) if there exists a neighborhood $U$ of $f$ in the $C^m$-topology such that every map $g \in U$ is topologically conjugate to $f$.
$f$ is structurally stable if $f$ is $C^1$-structurally stable.

Can someone give me a hint? Thank you!

• To prove that such diffeomorphism isn't structurally stable you have to provide a pair of diffeomorphisms in each neighbourhood of $A$ that for sure aren't conjugated. What if you consider only linear diffeomorphisms in these neighbourhoods? Can you find this pair of diffeomorphisms? – Evgeny Aug 23 '17 at 21:21
• I think that such a pair of diffeomorphisms could be the differential $DA_x$ for every $x$,but I don't know how to prove that it is not topologically conjugate to $A$. Can you give me more hints? Thank you! – g.pomegranate Aug 25 '17 at 10:22
• You have to play around the fixed point. There is everything known about behaviour near the fixed point of linear diffeomorphism. Few additional hints: do you know, how eigenvalues of diffeomorphism behave when you perturb it slightly? If you plot eigenvalues on complex plane and there is at least one at unit circle (i.e., non-hyperbolic diffeomorphism), how can they move on complex plane when you perturb diffeomorphism? Also, when two linear diffeomorphisms are conjugated to each other? – Evgeny Aug 25 '17 at 13:51
• I've been thinking about these hints, but I have not managed to solve the problem. Could you please help me to solve this ? Thank you! – g.pomegranate Aug 29 '17 at 9:07
• But did you manage to answer all of my questions? If you did, then in my opinion it puts everything in right places. – Evgeny Aug 30 '17 at 13:12