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I want to prove that the map $f: A \to A^{-1}$ is continuous where $A$ is an $n \times n$ matrix.

The idea is to use

$A^{-1}=\frac{1}{\det(A)}A_\text{adj}$ where $(A_\text{adj})_{ij}=(-1)^{i+j} \det(X_{ji}(A))$.

Here $X_{ji}(A)$ is the $(n-1) \times (n-1)$ matrix obtained by crossing out row $j$ and column $i$ from $A$.

I have already proven the continuity of the map $A \to a_{ij}$ $\forall i,j$, and thus that $A \to \det(A)$ is continuous as a composition of continuous maps.

Now it remains to show that the map $A \to A_\text{adj}$ is continuous. For this I need to show that the function $a_{ij} \to A_\text{adj}$ is continuous.

I know that the function $a_{ij} \to (A_\text{adj})_{ij}$ is continuous, but I cannot seem to deduce why this implies that $a_{ij} \to A_\text{adj}$ is continuous.

How do I prove this last step? I know there are some threads on this topic, but I could not find an answer specific to my question.

Thanks a lot!

Edit: I think I can use that for an $m \times n$ matrix,

$\|A\|_{operator} \leq mn \max_{i,j}|a_{ij}|$.

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  • $\begingroup$ That depends: with the definitions that you're working with, what does it mean for a function $g:\Bbb R \to \Bbb R^{m \times n}$ to be continuous? $\endgroup$ – Ben Grossmann Jun 24 '20 at 14:18
  • $\begingroup$ A minor is not a matrix. $\endgroup$ – Bernard Jun 24 '20 at 14:21
  • $\begingroup$ @Omnomnomnom Well, for matrices I would use the operator norm defined as the sup of Euclidean norm. $\endgroup$ – DerivativesGuy Jun 24 '20 at 14:27
  • $\begingroup$ @Bernard A minor is not a matrix, yes, but the adjugate is a matrix. $\endgroup$ – DerivativesGuy Jun 24 '20 at 14:28
  • $\begingroup$ Yes, of course. There remains a minor is a signed determinant, not a matrix. The phrasing may be confusing for beginners. $\endgroup$ – Bernard Jun 24 '20 at 14:31
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Claim: A map $f: \Bbb R \to \Bbb R^{m \times n}$ is continuous if all of the coordinate maps $f_{ij}:\Bbb R \to \Bbb R$ defined by $f_{ij}(t) = [f(t)]_{i,j}$ are continuous.

Proof: Note that the Frobenius norm, which is defined by $$ \|A\|_F^2 = \sum_{i,j}a_{ij}^2, $$ satisfies $\|A\| \leq \|A\|_F$ (where $\|A\|$ denotes the Euclidean operator norm). With that, we find the following. Consider any $x \in \Bbb R$. Fix $\epsilon > 0$. There exists a $\delta > 0$ such that if $|x-y| < \delta$, then $|f_{ij}(x) - f_{ij}(y)| < \epsilon/\sqrt{mn}$ for all $i,j$. With that, we find that $$ \|f(x) - f(y)\|^2 < \|f(x) - f(y)\|_F^2 = \sum_{ij} |f_{ij}(x) - f_{ij}(y)|^2 < \sum_{ij} \frac{\epsilon^2}{mn} = \epsilon^2. $$ So indeed, $f$ is continuous at $x$, which was arbitrary. So $f$ is continuous.

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  • $\begingroup$ Oh $x$ is fixed. My bad, sorry! $\endgroup$ – jijijojo Jun 24 '20 at 14:44
  • $\begingroup$ Thanks for your answer. I think I've already solved my problem (see edit). Could you check if that's fine as well? Thanks. $\endgroup$ – DerivativesGuy Jun 24 '20 at 14:58
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    $\begingroup$ @DerivativesGuy That should be $\max_{i,j}|a_{ij}|$, but otherwise that's fine. You could use your inequality in the same way that I used mine. $\endgroup$ – Ben Grossmann Jun 24 '20 at 15:11
  • $\begingroup$ That was a typo, good spot. Thanks a lot! $\endgroup$ – DerivativesGuy Jun 24 '20 at 15:27

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