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,

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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.