Invertible matrix perturbation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:00:58Z http://mathoverflow.net/feeds/question/94628 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94628/invertible-matrix-perturbation Invertible matrix perturbation Alex A 2012-04-20T12:08:45Z 2012-04-20T16:23:56Z <p>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). </p> <p>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 $x\in \mathbb{R} ^n$. </p> <p>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. </p> <p>Any ideas?</p> http://mathoverflow.net/questions/94628/invertible-matrix-perturbation/94650#94650 Answer by Michael Renardy for Invertible matrix perturbation Michael Renardy 2012-04-20T16:05:55Z 2012-04-20T16:05:55Z <p>Your idea works. Instead of trying to put A+B within distance 1 of the identity, just make t large and positive.</p> http://mathoverflow.net/questions/94628/invertible-matrix-perturbation/94651#94651 Answer by jerrysciencemath for Invertible matrix perturbation jerrysciencemath 2012-04-20T16:18:52Z 2012-04-20T16:23:56Z <p>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.</p>