invertibility of a matrix with a Gaussian perturbation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:56:24Z http://mathoverflow.net/feeds/question/110721 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110721/invertibility-of-a-matrix-with-a-gaussian-perturbation invertibility of a matrix with a Gaussian perturbation unknown (google) 2012-10-26T05:13:30Z 2013-01-31T11:22:00Z <p>Suppose that $A$ is an arbitrary fixed $n\times n$ matrix and $G$ a random $n\times n$ matrix with i.i.d. $N(0,1)$ entries. Is there a simple proof that $A+G$ is invertible with probability 1?</p> <p>What if $G$ is a random Wigner matrix (symmetric, upper diagonal entries are i.i.d. $N(0,1)$)? Is $A+G$ still invertible with probability 1? Is there a simple proof?</p> http://mathoverflow.net/questions/110721/invertibility-of-a-matrix-with-a-gaussian-perturbation/119052#119052 Answer by Davide Giraudo for invertibility of a matrix with a Gaussian perturbation Davide Giraudo 2013-01-16T10:47:15Z 2013-01-17T10:24:59Z <p>We use the idea suggested by <a href="http://mathoverflow.net/users/1590/alekk" rel="nofollow">Alekk</a>.</p> <p>Let $A_{i,j}$ the entries of $A$. Then in the first case, the entries of $M:=A+G$ are Gaussian independent random variables, that is, $M_{i,j}\sim N(A_{i,j},1)$. Denote $N$ the set of elements of $\Bbb R^{n^2}$ such that the matrix of generic term $x_{i,j}$ is not invertible. This set has null Lebesgue measure in $\Bbb R^{n^2}$ as it's the zeros of a polynomial. By independence, the family $(M_{i,j},i,j\in [n])$ is Gaussian, so $$p:=P(A+G\mbox{ is not invertible})=P((M_{i,j})_{i,j=1}^n\in S).$$</p> <p>As the law of $(M_{i,j},i,j\in [n])$ is absolutely continuous with respect to Lebesgue measure in $\Bbb R^{n^2}$, we conclude that $p=0$. </p> <p>In the second case, write $\det(A+G)$ as a polynomial of the $G_{i,j},i\leqslant j$, and use the fact that $(G_{i,j},1\leqslant i\leqslant j\leqslant n)$ is Gaussian to conclude in the same way as in the first case.</p> <ul> <li>The result doesn't depend on the choice of the <em>deterministic</em> matrix $A$.</li> <li>We don't need i.i.d.ness, just the fact that $(G_{i,j},i,j\in [n])$ is Gaussian in the first case, $(G_{i,j},i\leqslant j,i,j\in [n])$ in the second.</li> <li>We can have a more general result when the law of $(A_{i,j}+G_{i,j})_{i,j\in [n]}$ is absolutely continuous with respect to Lebesgue measure on $\Bbb R^{n^2}$.</li> </ul>