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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?

flag	asked Apr 20 2012 at 12:08 Alex A 274●2●8

	What does it mean for $B$ to be supported away from $x_0?$ – Igor Rivin Apr 20 2012 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 $\{x: B(x)\neq 0 \}$. – Alex A Apr 20 2012 at 14:33

jerrysciencemath · Accepted Answer · 2012-04-20 16:23:56Z UTC

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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.

edited Apr 20 2012 at 16:23

answered Apr 20 2012 at 16:18

jerrysciencemath
577●1●12

Thanks a lot! Your solution appears to work very well. – Alex A Apr 21 2012 at 9:57

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 $x_0$? – Alex A Apr 28 2012 at 20:13

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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 $B(x)=\overline{B}(x)A(x_0)$. – jerrysciencemath Apr 29 2012 at 9:16

Of course, so obvious once you see it! Thanks! – Alex A Apr 30 2012 at 13:57

Michael Renardy · Answer 2 · 2012-04-20 16:05:55Z UTC

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Your idea works. Instead of trying to put A+B within distance 1 of the identity, just make t large and positive.

answered Apr 20 2012 at 16:05

Michael Renardy
4,771●1●10●17

Yes, I realized this. Thanks! – Alex A Apr 21 2012 at 9:58

Invertible matrix perturbation

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