# adjugate $A_\text{adj}$ is a continuous function of the matrix $A$

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,

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

• 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? – Ben Grossmann Jun 24 '20 at 14:18
• A minor is not a matrix. – Bernard Jun 24 '20 at 14:21
• @Omnomnomnom Well, for matrices I would use the operator norm defined as the sup of Euclidean norm. – DerivativesGuy Jun 24 '20 at 14:27
• @Bernard A minor is not a matrix, yes, but the adjugate is a matrix. – DerivativesGuy Jun 24 '20 at 14:28
• Yes, of course. There remains a minor is a signed determinant, not a matrix. The phrasing may be confusing for beginners. – Bernard Jun 24 '20 at 14:31

## 1 Answer

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.

• Oh $x$ is fixed. My bad, sorry! – jijijojo Jun 24 '20 at 14:44
• Thanks for your answer. I think I've already solved my problem (see edit). Could you check if that's fine as well? Thanks. – DerivativesGuy Jun 24 '20 at 14:58
• @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. – Ben Grossmann Jun 24 '20 at 15:11
• That was a typo, good spot. Thanks a lot! – DerivativesGuy Jun 24 '20 at 15:27