I have a second rank tensor $\mu$, which is in my case a linear map from $S^3(\mathbb{R})$ to $S^3(\mathbb{R})$ where $S^3(\mathbb{R})$ are the $3 \times 3$ symetric matrices with real coefficients.
I was wondering if there exists a somehow "natural way" to define the "inverse tensor" of $\mu$. As an example, maybe I could state something like "$\mu$ is inversible if $\mu (E)$ is inversible for any $E \in S^3(\mathbb{R})$, and we pose $\mu^{-1}(E)=(\mu(E))^{-1}$"
Do you guys have some thoughts on this ? Any ideas are welcomed !
Possible answer :
As $\mu$ is a linear map from $E=S³(\mathbb{R})$ to $E$, it can be seen as a matrice $U$ of dimension $6 \times 6$ by fixing a base $\mathcal{B}$ of $E$. Then I could just say that $\mu$ is inversible if the matrice $U$ is, and set $\mu^{-1}$ the application from $E$ to $E$ such as it's matrice in the base $\mathcal{B}$ is $U^{-1}$. This definition seems very natural since it verifies
$$\mu \circ \mu^{-1} = Id_{E \rightarrow E}.$$