Let $A$ be an $n\times n$ matrix which depends smoothly on a variable $x\in \mathbb{R}^n$ and such that there are constants $C\_\alpha > 0 $ so that $\| \partial ^\alpha A \| \le C\_\alpha $ for all multi-indices $\alpha \in \mathbb{N}\_0^n$ (i.e. $A$ together with all its derivatives are bounded in matrix norm). Call the set of such matrices $S$. Assume moreover that at some point $x\_0 \in \mathbb{R}^n$ we have $A(x\_0) = I$ (identity matrix). 


I now want to prove that we can find a matrix $B\in S$ that is supported away from $x\_0$ and such that $A + B$ is invertible for all $x\in \mathbb{R} ^n$. 


My approach was that, since $A$ is smooth, we can find neighborhoods $V\_0$ and $V\_1$ of $x\_0$ with $\overline{V\_0}\subset V\_1$ such that 
$$
\|A - I\| \le 1/3 \text{ on } V\_0 \quad \text{ and } \quad 
\|A - I\| \le 2/3 \text{ on } V\_1
$$
and then take $B=t\chi I$ for some suitable $t\in \mathbb{R}$ where $\chi $ is smooth and $\chi = 0$ on $V\_0$ and $\chi = 1$ outside $V\_1$. I tried making $A+B$ within distance 1 to identity but did not succeed. 


Any ideas?



Invertible matrix perturbation
Define $f(x)$ to be the least real eigenvalues of $A(x)$, then $f$ is defined in a neighborhood $B(x\_0,\epsilon)$ of $x\_0$ and $f(x)>1/2$ for any $x\in B(x\_0,\epsilon)$. Suppose $M>\sup\_x||A(x)||$, then choose $\chi(x)$ to be a nonnegative function such that $\chi$ is smooth, $\chi(x)=0$ for any $x\in B(x\_0,\epsilon/2)$, and $\chi(x)=M$ for any $x\notin B(x\_0,\epsilon)$. Put $B(x)=\chi(x)I$, then $B$ satisfies your requirement, since all real eigenvalues of $A+B$ is always positive, hence $A+B$ is always invertible.


