Let $M:\mathbb{R}^n\rightarrow GL(n,\mathbb{R})$ be such that for some $C>0$
$\frac{1}{C}\vert x\vert^2\leq x^TM(.)x\leq C\vert x\vert^2$
for all $x\in\mathbb{R}^n$. Does the map $x\mapsto M(x)^{-1}$ satisfy a similar property?
My idea is to look at
$\frac{1}{C}\vert M(y)^{-1}x\vert^2\leq x^T M(y)^{-1} x\leq C\vert M(y)^{-1}x\vert^2$.
I think that there should be a way to show that the rhs $\leq C\vert x\vert^2$ but I do now know how to determine the operator norm of $M(y)^{-1}$... The lhs I do not know how to bound from below so that the $y$ does not appear there any more...
The lower bound for $M$ implies upper bound for $M^{-1}$, as follows. Since $x^T Mx\ge |x|^2/C$, it follows that $|Mx|\ge |x |/C$ for all $x$. Hence $\|M^{-1}\|\le C$, which in turn yields $x^TM^{-1}x\le C|x|^2$.
But the lower bound for $M^{-1}$ need not hold. Consider $M:\mathbb R^2\to GL(2,\mathbb R)$ defined by $$M(y_1,y_2) = \begin{pmatrix} 1 & y_1 \\ -y_1 & 1 \end{pmatrix} $$ Observe that $x^TMx=|x|^2$ for all $x\in\mathbb R^2$. But $$M(y_1,y_2)^{-1} = \frac{1}{y_1^2+1}\begin{pmatrix} 1 & y_1 \\ -y_1 & 1 \end{pmatrix} $$ hence $x^TMx=|x|^2/(y_1^2+1)$ for all $x\in\mathbb R^2$. Since $y_1$ can be arbitrarily large, the lower bound fails.
If $M$ is assumed to be symmetric, you do get both bounds, from the consideration of the eigenvalues of $M$.