# Invertible matrices of type $A+\varepsilon I$ [closed]

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?

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

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• What is assumed about $A$ by "some authors"? – kimchi lover Mar 10 at 0:00
• Is $A$ from $GL_n(\mathbb{C})$? – Minus One-Twelfth Mar 10 at 0:00
• 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. – stefano Mar 10 at 7:39

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 $$0<|t|<\mu$$.