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Let $\{A_k\}_{k=1}^{\infty}$ be a sequence of real $n \times n$ matrices. Suppose $A_k$ is invertible for all $k \in \mathbb{N}$, that $\lim_{k \rightarrow \infty} A_k = A$, and that $A$ is invertible. Is $\lim_{k \rightarrow \infty} A_k^{-1} = A^{-1}$?

This feels like it should be true but I'm having trouble coming up with a proof.

Also, I think defining $\lim_{k \rightarrow \infty} A_k = A$ as componentwise convergence makes the most sense for my problem but any other insights are welcome!

Brian Lai
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    What is $A_k$ ? a number ? a matrix ? a function ? If it's a matrix (as the tags let me guess), how do you define $\lim_{k\to \infty }A_k=A$ ? componentwise ? – Surb Mar 11 '21 at 20:30
  • Good point, I suppose componentwise makes the most sense for my context – Brian Lai Mar 11 '21 at 20:35
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    All norms on finite-dimensional vector spaces are equivalent, and component-wise convergence is equivalent to convergence in any of them. – saulspatz Mar 11 '21 at 20:44
  • @saulspatz: This doesn't answer the question : "how does the OP defines $\lim_{k\to \infty }A_k=A$ ?". – Surb Mar 11 '21 at 21:05
  • @Surb: I think the edit to the original post, defining convergence componentwise, is adequate (though if you wanted to explore alternative meanings in an Answer, that is invited). – hardmath Mar 12 '21 at 02:44
  • I think it likely that this has been asked and answered before. It follows from the continuity of matrix inversion with a bit of bookkeeping for the sequence of matrices. – hardmath Mar 12 '21 at 02:47

2 Answers2

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Sine $A_k$ converge, so do their determinants (which are polynomial functions of the entries), and the limit of the determinant is NOT zero, so do their cofactors (which are polynomial functions of the entries) and so do the cofactors over the determinant, which are the entries of the inverses.

Igor Rivin
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Your question is asking whether $\Phi : GL(n, \mathbb{R}) \to GL(n, \mathbb{R})$ defined by $$\Phi(A) = A^{-1}$$ is continuous. This is true. In fact, $\Phi$ is differentiable. Let $||\cdot||$ denote the operator norm on $M(n, \mathbb{R})$. A geometric series type argument yields that for $H \in M(n, \mathbb{R})$ with $||H|| < 1$, \begin{align} (I + H)^{-1} &= \sum_{k = 0}^{\infty}(-1)^kH^k \\ &= I - H + H^2 - \dots \\ &= I - H + O(||H||^2). \end{align} Now let $A \in GL(n, \mathbb{R})$ be arbitrary. For $H \in M(n, \mathbb{R})$ we have $$(A + H) = A(I + A^{-1}H)$$ Hence if $||A^{-1}H|| < 1$, then $A + H$ is invertible. Since $||A^{-1}H|| \leq ||A^{-1}|| \cdot ||H||$, $A + H$ is invertible whenever $||H|| < \frac{1}{||A^{-1}||}$ and \begin{align} (A + H)^{-1} &= (A(I + A^{-1}H))^{-1} \\ &= (I + A^{-1}H)^{-1}A^{-1} \\ &= (I - A^{-1}H + O(||H||^2))A^{-1} \\ &= A^{-1} - A^{-1}HA^{-1} + O(||H||^2). \end{align} Hence $D\Phi(A)H = -A^{-1}HA^{-1}$.

Mason
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