Differentiability of a map from $GL(n)$ into itself

Question

Let $GL(n):=\{M\in \mathbb{R}^{n\text{ x }n}\mid \text{det(M)}\neq 0\}$. For an arbitrary natural number $k$, prove that the map $\Psi_k:GL(n) \to GL(n)$ defined by $$\Psi_k(M) = (M^{-1})^{k}$$

is differentiable.

I don't feel comfortable with this problem, since I'm not used to talk about differentiability of functions envolving matrices. Actually, I don't even know wherer to start.

(by the way, I believe we're supposed to use the matrix norm $||M||=\sup_{|v|= 1}|M(v)|$ (where $v \in \mathbb{R}^n$ and |.| is the usual euclidean norm), since this has been the standard in my analysis course)

Answer 1

Split your issue into proving that:

• taking the inverse $A \rightarrow A^{-1}=B$ is differentiable (http://planetmath.org/derivativeofinversematrix), and

• taking a matrix at the power $n$: $B \rightarrow B^n=C$ is differentiable (Is there a general form for the derivative of a matrix to a power?).

Then the composite operation $A \rightarrow B \rightarrow C$ is differentiable.