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 enter image description here

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.

  • $\begingroup$ Didn't you switch $A$ and $B$ in $D_{A} \mbox{inv}(B)$? $\endgroup$ – Rodrigo de Azevedo Mar 16 at 15:18
  • $\begingroup$ @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. $\endgroup$ – Viktor Glombik Mar 16 at 15:19
  • $\begingroup$ 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. $\endgroup$ – Rodrigo de Azevedo Mar 16 at 15:22
  • $\begingroup$ $\mbox{inv}(B)$ is a matrix, not a function. It has no derivative. $\endgroup$ – Rodrigo de Azevedo Mar 16 at 15:23
  • $\begingroup$ @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 $f$. $\endgroup$ – Viktor Glombik Mar 16 at 15:28

Your Answer

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

Browse other questions tagged or ask your own question.