(How) can one visualise the derivative of the function $A \mapsto A^{-1}$, where $A$ is a matrix?

In class, we have shown:

Let $$V$$ be a finite-dimensional Banach-space, then the general linear group $$\mbox{GL}(V) := \{ A \in L(V, V): A \text{ is invertible } \}$$ is open in $$L(V,V)$$ and the mapping $$\mbox{inv}: \mbox{GL}(V) \to L(V,V), \quad A \mapsto A^{-1}$$ is differentiable and its derivative is $$D_{A} \mbox{inv}(B) = - A^{-1} B A^{-1}$$.

This is the case because $$\mbox{inv}(B) = \mbox{inv}(A) \underbrace{- A^{-1}(B - A)A^{-1}}_{D_A \text{inv}(B - A)} + \underbrace{A^{-1}(B - A)(A^{-1} - B^{-1})}_{= R(B)}.$$

Is there any way to visualise what $$\mbox{inv}$$ looks like and to picture its derivative?

Any help is greatly appreciated.

Using @RodrigodeAzevedo's hint, the vector field of a the inverse of a symmetric $$2 \times 2$$ matrix looks like this

Our definition of differentiable is:

Let $$V$$ and $$W$$ be Banachspaces, $$V$$ finite dimensional, and $$G \subset V$$ an open subset. We call a function $$f: G \to W$$ differentiable in $$p \in G$$ if there exists a linear map $$F: V \to W$$, so that for the remainder function $$R:G \to W$$, defined by $$f(x) = f(p) + F(x - p) + R(x)$$ we have $$\frac{R(x)}{\| x - p \|} \xrightarrow{x \to p} 0$$.

Then, the function $$F$$ is unique and called the differential of $$f$$ in $$p$$, we write $$F = D_p f$$.

Lemma: With all names from above we have for all $$v \in V$$ $$F(v) = \lim_{t \to 0} \frac{f(p + tv) - f(p)}{t} =: \partial_v f(p),$$ if the limit exists and call it the directional derivative.

• Didn't you switch $A$ and $B$ in $D_{A} \mbox{inv}(B)$? – Rodrigo de Azevedo Mar 16 at 15:18
• @RodrigodeAzevedo Nope. It's the derivative of inv(B) in the "point" A. Notice $D_p f(v) = \partial_v f(p)$ in out notation. – Viktor Glombik Mar 16 at 15:19
• I believe you did. You have the directional derivative of $\mbox{inv}$ at $A$ in the direction of $B$. Or, $p = A$, $v = B$. Now check your notation again. – Rodrigo de Azevedo Mar 16 at 15:22
• $\mbox{inv}(B)$ is a matrix, not a function. It has no derivative. – Rodrigo de Azevedo Mar 16 at 15:23
• @RodrigodeAzevedo Well, a matrix is a linear map and therefore it's one derivative, right? Since we have $f(x) = f(p) + f(x - p) + 0$ for any linear map $f$. – Viktor Glombik Mar 16 at 15:28