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It is well known that $\det$ function is continuous and, by $$ \det.^{-1}(\mathbb{C}^{*})=GL_n(\mathbb{C}),$$ the set $GL_n(\mathbb{C})$ is a open subset of $M_n(\mathbb{C})$. From here, some authors, claim that the matrices $$ A+\varepsilon I,\quad \forall \space 0<\varepsilon<\mu$$ are invertible. Why is that true?

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closed as off-topic by Eevee Trainer, Shailesh, Alexander Gruber May 11 at 5:27

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    $\begingroup$ What is assumed about $A$ by "some authors"? $\endgroup$ – kimchi lover Mar 10 at 0:00
  • $\begingroup$ Is $A$ from $GL_n(\mathbb{C})$? $\endgroup$ – Minus One-Twelfth Mar 10 at 0:00
  • $\begingroup$ This analysis trick was used to show some formula for the case when the matrix $A$ is not invertible. For instance, $adj(AB)=adj(A)adj(B)$, for all matrices $A,B\in M_n(\mathbb{C})$. Some authors means, for example, Fuzhen Zhang or Titu Andreescu. $\endgroup$ – stefano Mar 10 at 7:39
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Since the polynomial $p(t)=\det(tI+A)$ has at most $n$ roots, all roots of $p(t)$ are isolated. So there is $\mu>0$ such that $p(t)\ne 0$ for all $0<|t|<\mu$, implying that $A+tI$ is invertible for all $0<|t|<\mu$.

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