Properties of matrix $T_\epsilon$ such that there exists a unique inverse for $\epsilon \in (-\delta,\delta)$

Question

I'm considering an $n\times n$ matrix $T_\epsilon$ such that there exists a unique inverse for $\epsilon \in (-\delta,\delta)$ for some $\delta > 0$. I'm trying to determine several quantities of interest:

$$\frac{d}{d\epsilon} T_\epsilon^{-1} \tag{1}$$ $$\frac{d}{d\epsilon} \det T_\epsilon \tag{2}$$ $$\frac{d}{d\epsilon} \textrm{cof} T_\epsilon \tag{3}$$ $$\frac{d}{d\epsilon} \textrm{tr} (T_\epsilon^2) \tag{4}$$ $$\frac{d}{d\epsilon} \textrm{tr} ((\textrm{dev} T_\epsilon)^2) \tag{5}$$

What is $T_\epsilon$ even though?? Is this $T+\epsilon$? Or $T+\epsilon I$? Afterwards I think it's a matter of using things like the cofactor/adjugate, determinant, Jacobi's theorem, etc. but I want to be clear about the matrix first. Do y'all have any ideas? Thank you.

EDIT Okay so thanks to u/Delta-u, if we're thinking about $T(\epsilon)\rightarrow T_\epsilon$, I saw somebody use this trick: \begin{align*} T_\epsilon T_\epsilon^{-1} &= I \\ \frac{d}{d\epsilon} (T_\epsilon T_\epsilon^{-1}) &= 0 \\ \frac{dT_\epsilon}{d\epsilon} T_\epsilon^{-1} + T_\epsilon\frac{dT_\epsilon^{-1}}{d\epsilon} &= 0 \\ \frac{dT_\epsilon^{-1}}{d\epsilon} &= T_\epsilon^{-1}\frac{dT_\epsilon}{d\epsilon} T_\epsilon^{-1} \end{align*} Does this seem like the right track? I can keep going with this then

Answer 1

Hint

In my opinions the main ideas are:

• Use the definition of the derivative as an order $1$ expansion.
• Reduce the problem to the derivative around $I$ if possible.

For the first derivative using these two ideas:

With $\epsilon \in (-\delta,\delta)$ and $h$ such that $e=\epsilon+h \in(-\delta,\delta)$ you have:

$$T_{\epsilon+h}=T_\epsilon+\frac{dT_\epsilon}{d \epsilon} h+o(h)$$ so: $$T_{\epsilon+h}^{-1}=\left(T_\epsilon+\frac{dT_\epsilon}{d \epsilon} h+o(h) \right)^{-1}=T_\epsilon^{-1} \left(I +\frac{dT_\epsilon}{d \epsilon}T_\epsilon^{-1} h+o(h) \right)^{-1}$$ but around the identity you have: $$(I+A)^{-1}=I-A +o(A)$$ so: $$T_{\epsilon+h}^{-1}=T_\epsilon^{-1} \left(I - \frac{dT_\epsilon}{d \epsilon}T_\epsilon^{-1} h+o(h) \right)=T_\epsilon^{-1}-T_\epsilon^{-1} \frac{d T_\epsilon}{d \epsilon}T_\epsilon^{-1}h +o(h)$$

so: $$\frac{d T_\epsilon^{-1}}{d \epsilon}=-T_\epsilon^{-1} \frac{d T_\epsilon}{d \epsilon}T_\epsilon^{-1}$$