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Suppose that $A,A_1,A_2,\ldots\in\mathbb R^{p\times p}$ are invertible square matrices such that $\|A-A_n\|=o(a_n)$ as $n\to\infty$, where $a_n\to0$ as $n\to\infty$.

Is it true that $\|A^{-1}-A_n^{-1}\|=o(a_n)$ as $n\to\infty$ as well?

Here is my attempt. We have that \begin{align*} A^{-1}-A_n^{-1} &=A^{-1}(A_n-A)A_n^{-1}\\ &=A^{-1}(A_n-A)A_n^{-1}AA^{-1}\\ &=A^{-1}(A_n-A)(I-A^{-1}[A-A_n])^{-1}A^{-1}. \end{align*} Using the properties of the Frobenius norm (see here), we obtain \begin{align*} \|A^{-1}-A_n^{-1}\| &\le\|A^{-1}\|^2\|A_n-A\|\|(I-A^{-1}[A-A_n])^{-1}\|. \end{align*} Since $\|A^{-1}[A-A_n]\|<1$ for sufficiently large values of $n$, we have that $$ \|(I-A^{-1}[A-A_n])^{-1}\| \le\sum_{k\ge0}\|(A^{-1}[A-A_n])^k\| \le\sum_{k\ge0}(\|A^{-1}\|\|A-A_n\|)^k<\infty $$ for sufficiently large values of $n$. We conclude that $\|A^{-1}-A_n^{-1}\|=o(a_n)$ as $n\to\infty$.

Is this correct? Is there a simpler way to show this? The only not straightforward part of the proof is to show that $\|A_n^{-1}\|$ is bounded. Is there an easy to way to see that $\|A_n^{-1}\|$ is bounded?

Any help is much appreciated.

Cm7F7Bb
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1 Answers1

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The map $B \to B^{-1}$ is continuous on the invertible $n \times n$ matrices, using the formula $B^{-1} = \text{adj}(B)/\det(B)$, where $\text{adj}(B)$ is the adjugate of $B$. Since $\{A_n\}_{n=1}^\infty \cup \{A\}$ is a compact set of invertible matrices, so is $\{A_n^{-1}\}_{n=1}^\infty \cup \{A^{-1}\}$. Therefore it is bounded.

Robert Israel
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