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?


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?



invertibility of a matrix with a Gaussian perturbation
We use the idea suggested by [Alekk](https://mathoverflow.net/users/1590/alekk).


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 
$$p:=P(A+G\mbox{ is not invertible})=P((M\_{i,j})\_{i,j=1}^n\in S).$$


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 $p=0$. 


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.


* The result doesn't depend on the choice of the *deterministic* matrix $A$.
* 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.
* 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 $\Bbb R^{n^2}$.


