# Why the addition of the noise will almost certainly make the square matrix invertible?

Why the addition of the noise will almost certainly make the square matrix invertible?

The matrix is invertible if its determinant is not equal to $$0$$. The determinant of the matrix geometrically can be understood as a $$n$$-dimensional parallelepiped defined by its row/column vectors.
Now let our non-invertible matrix $$A$$ be for simplicity $$3 \times 3$$ matrix. Obviously, $$det(A) = 0$$ i.e. the parallelepiped formed by the rows/columns of $$A$$ is flat. If we add some noise matrix $$V$$ where $$v_{ij}$$~$$Norm(0,a)$$ to each entry of $$A$$ i.e. $$A+V$$, we will change the the rows/columns of $$A$$. How can one show geometrically (or any way) that the new rows/columns would be independent?

Denote by $$V$$ your random matrix of $$v_{i,j}$$'s. Note that $$\textrm{det}(A+V)$$ is a polynomial in the entries $$V$$.
Appealing to the implicit function theorem, the level sets of $$B\mapsto \textrm{det}(A+B)$$ are co-dimension $$1$$ manifolds (in $$\mathbb{R}^{n^2}$$) and hence, are Lesbegue null-sets. Since $$V$$ is Gaussian and hence, is absolutely continuous with respect to the Lesbegue measure, for any $$x\in \mathbb{R}$$, we have that $$\mathbb{P}(\textrm{det}(A+V)=x)=0$$. This holds, in particular, for $$x=0$$ and we conclude that $$A+V$$ is almost surely invertible.