Choosing a smooth diagonalizable approximation of a smooth map into $\text{Aut}(\mathbb R^{n}) = \textbf{GL}(\mathbb R^{n}) \subset M_{n}(\mathbb R)$

Question

Suppose we have a map $\mathbb R^n \to \text{Aut}(\mathbb R^{n}) = \textbf{GL}(\mathbb R^{n})$ with $A=(A_{ij}(x))$ and $A_{ij}$ all being smooth.

It is known that diagonalizable matrices with complex values are dense in set of $n\times n$ complex matrices, in particular matrices with real entries is dense in $M_n(\mathbb C)$ by the same proof if we allow diagonalization over $\mathbb C$. Thus for every $x\in \mathbb R^n$, we can choose a diagonalizable matrix $B_\varepsilon(x)$ with $\|A(x)-B_\varepsilon(x)\|<\varepsilon$.

However, fixing a compact set $K\subseteq \mathbb R^n$, can we choose a map $x\mapsto B_\varepsilon(x)$ wih ${(B_\varepsilon)}_{ij}$ smooth such that $\|A(x)-B_\varepsilon(x)\|<\varepsilon$ holds for all $x \in K$?

Answer 1

Not an answer, but too long for a comment. Your statement "in particular matrices with real entries is dense in $M_n(\mathbb C)$ by the same proof if we allow diagonalization over $\mathbb C$." does not hold because of the trace $$ tr:\ M_n(\mathbb C)\to \mathbb C $$ which is invariant by conjugation and smooth.
Proof If $M\in M_n(\mathbb R)$ then $tr(PMP^{-1})=tr(M)\in \mathbb R$ which is closed in $\mathbb C$, then $V=tr^{-1}(\mathbb R)$ is closed, different from the whole $M_n(\mathbb C)$ (does not contain $iI_{n\times n}$ for instance) and closed by conjugation.
Hope it helps.