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Problem Prove that the inverse operator $\text{Inv}:A \mapsto A^{-1}$ (where $A$ is an $n\times n$ invertible matrix) is continuous

Context : I'm trying to do the problem $5.32$ (Pugh, 2nd). The alternative $(c)$ asks to show that the inverse operator $\text{Inv}:A \mapsto A^{-1}$ is a homeomorphism $G$ onto $G$.

Question 1 is the following approach correct ?

For a given $\epsilon>0$ we have that

\begin{align} \rvert\rvert Y^{-1} - X^{-1} \rvert\rvert &=\; \rvert\rvert Y^{-1}\;(Id-YX^{-1}) \rvert\rvert \\ &\leq\; \rvert\rvert Y^{-1} \rvert\rvert \; \rvert\rvert Id-Y X^{-1}\rvert\rvert \\ &=\; \rvert\rvert Y^{-1} \rvert\rvert \; \rvert\rvert (X-Y)X^{-1} \rvert\rvert \\ &\leq\; \rvert\rvert Y^{-1} \rvert\rvert \; \rvert\rvert X^{-1} \rvert\rvert \; \rvert\rvert X-Y \rvert\rvert \end{align}

Thus we just have to choose $\rvert\rvert X-Y \rvert\rvert < \frac{\epsilon}{\rvert\rvert Y^{-1} \rvert\rvert \; \rvert\rvert X^{-1} \rvert\rvert}$ which goes to zero because both $Y^{-1}, X^{-1}$ are invertible matrices implying they never vanish.

Powder
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    This is not correct. You have to know what happens to the denominator when $Y \to X$. – Kavi Rama Murthy Jul 06 '21 at 23:14
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    The inverse is only continuous if you restrict to invertible matrices. That’s important. You still need to show that everything in a neighborhood of an invertible matrix is invertible and that $||Y^{-1}||$ is bounded in some neighborhood of $X$. – Eric Jul 06 '21 at 23:50
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    Your second approach is wrong. There is no inverse determinant function. – Eric Jul 06 '21 at 23:50
  • Thank you for those tips I'll think about that. (For the second one, I think this shows me that I need to take some rest rs) – Powder Jul 06 '21 at 23:54
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    Relevant: https://math.stackexchange.com/questions/810670/continuity-of-the-inverse-matrix-function – Clement C. Jul 07 '21 at 00:03
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    Show that every step of this process is continuous: https://www.mathsisfun.com/algebra/matrix-inverse-minors-cofactors-adjugate.html – David Lui Jul 07 '21 at 00:06
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    One nice approach to show that $GL(n, \mathbb{R})$ is open and that the map $\Phi \colon GL(n, \mathbb{R}) \to GL(n, \mathbb{R})$ defined by $\Phi(A) = A^{-1}$ is differentiable is to use the "geometric series" $$(I - A)^{-1} = \sum_{n = 0}^{\infty}A^n,$$ absolutely convergent for $||A|| < 1$. – Mason Jul 07 '21 at 01:59
  • Related: https://math.stackexchange.com/q/3138681/169085 – Alp Uzman May 15 '22 at 08:23
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    @KaviRamaMurthy can you elaborate? I am looking at this person's attempt and I don't understand why it is wrong. Are you referring to the map $XY^{-1}$? – Lemon Nov 10 '22 at 08:24

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