# Invertible matrix perturbation

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?

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What does it mean for $B$ to be supported away from $x_0?$ – Igor Rivin Apr 20 '12 at 14:12
All I want is that $x_0$ does not belong to the support of $B$ where the support of $B$ is the closure of $\{x: B(x)\neq 0 \}$. – flavio Apr 20 '12 at 14:33

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.
Is there a similar argument if A is merely assumed to be invertible at the point $x_0$, so there is also the possibility of one eigenvalue being negative at $x_0$? – flavio Apr 28 '12 at 20:13
Apply the proof to $\overline{A}(x)=A(x)(A(x_0))^{-1}$, since we have $\overline{A}(x_0)=I$, we can always find a $\overline{B}(x)$ such that $\overline{A}(x)+\overline{B}(x)$ is invertible, then put $B(x)=\overline{B}(x)A(x_0)$. – Yuchen Liu Apr 29 '12 at 9:16