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Suppose that $A_1 , A_2 \in M_n (\mathbb C)$ are invertible matrices.
Claim: There exists finitely many complex numbers $z\in \mathbb C$ such that $f(t) :=\det(zA_1+ (1-z)A_2)=0.$ Which is true via Fundamental theorem of algebra. Now define $$\gamma:[0,1]\longrightarrow \mathbb C.$$
Would it easily follow by the construction above that there exists a continuous path $\varphi(z)$ of the form $$\varphi(z)= \gamma(z) A_1 +(1-\gamma(z))A_2$$ connecting $A_1$ to $A_2$?

Any help and hints would be so much appreciated!

Tokita Ohma
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    Did you mean to write $(1 - γ(z)) A_2$ instead of the current $(1 - γ(z) A_2)$ in the formula vor $φ$? – Jendrik Stelzner Dec 13 '22 at 11:54
  • The answers on this question should help. – Bruno B Dec 13 '22 at 12:07
  • @JendrikStelzner yes sorry for the typo error. – Tokita Ohma Dec 13 '22 at 14:18
  • @BrunoB Thanks! I don't understand the argument, we can find a path $\gamma\colon[0,1]\to\mathbb C$ is this immediate? – Tokita Ohma Dec 13 '22 at 14:19
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    @TokitaOhma See the accepted answer on the post. To reframe the argument into your context: the set of $t$ for which $f(t) = 0$ is finite, and it follows that there is a path $\gamma:[0,1] \to \Bbb C$ from $1$ to $0$ in $\Bbb C$ such that $f(\gamma(t)) \neq 0$ for all $t \in [0,1]$. – Ben Grossmann Dec 13 '22 at 17:10
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    I think you mean $A_1 , A_2 \in GL_n (\mathbb C)$. Everything in $M_n (\mathbb C)$ is trivially path connected (e.g. to the zero matrix). – user8675309 Dec 13 '22 at 17:16
  • @BenGrossmann sorry I still can't reason out why would it follow that there is a path $\gamma:[0,1] \to \Bbb C$ is this because $f(t)=0$ is finite? Would finiteness of $0$ would easily follow the existence of path connectedness? – Tokita Ohma Dec 13 '22 at 22:52
  • The standard approach in a complex analysis text: consider some finite set $S$. Then it's immediate that $\mathbb C-S$ is open; prove it is connected. Now prove (/ use the fact) that open and connected implies path connected in a metric space (locally path connected). It's worthwhile to compare this with $\mathbb R$. – user8675309 Dec 13 '22 at 23:46
  • @Tokita It follows from the fact that the set of $z \in \Bbb C$ for which $f(z) = 0$ is finite. I'm not sure what you have in mind when you say "$f(t) = 0$ is finite" or "finiteness of $0$". – Ben Grossmann Dec 14 '22 at 03:57

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