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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 $x \in K$?

Markus Klyver
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  • So A maps from R^n to GL(R^n) and both A and its coordinates in a chosen basis are smooth with respect to the standard topology? And A is fixed in the last part? – saolof Mar 25 '20 at 23:43
  • @saolof Yes, in the induced topology given by the inclusion $\textbf{GL}(\mathbb R^n) \hookrightarrow \mathbb R^{n^2}$.

    And $x\mapsto A(x)$ is indeed a fixed map $\mathbb R^n\to\textbf{GL}(\mathbb R^n)$.

     – Markus Klyver Mar 25 '20 at 23:48
  • R is not algebraically closed though, so in general $B_{\epsilon}$ will be complex? – saolof Mar 25 '20 at 23:53
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    @saolof Yeah, that's what I meant by allowing diagonalization over $\mathbb C$. – Markus Klyver Mar 25 '20 at 23:57
  • If the roots of a complex polynomial depend smoothly on the coefficients, I think this wouldn't be so hard to prove, but I don't know whether that's true. – Camilo Arosemena Serrato Mar 26 '20 at 00:23

1 Answers1

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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.

Duchamp Gérard H. E.
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