Does $(\mathbf A+\epsilon \mathbf I)^{-1}$ always exist? Why? Does $(\mathbf A+\epsilon \mathbf I)^{-1}$ always exist, given that $\mathbf A$ is a square and positive (and possibly singular) matrix and $\epsilon$ is a small positive number? I want to use this to regularize a sample covariance matrix ($\mathbf A = \Sigma$) in practice, so that I can compute the inverse, which I need to calculate a Mahalanobis distance between two samples. In practice, my covariance matrix is often singular. I know the term $(\mathbf A+\epsilon \mathbf I)^{-1}$ often appears in the context of least squares problems involving Tikhonov regularization (ridge regression). However, I've never seen a statement, proof, or reference which says that the expression is always invertible.
Can any of you help me with a proof or reference?

Because the determinant function is continuous, and because $A$ is invertible if and only if $\det A \neq 0$, we have that $\det(A)\neq 0 \Rightarrow \det(A+\epsilon B)\neq 0$ for sufficiently small $\epsilon$, regardless of what $B$ is.
If $A$ is singular, then we note that the characteristic polynomial of $A$, given by $\operatorname{ch}_A(t)=\det(A-tI)$ is a polynomial in $t$, and therefore has finitely many roots.  Then, whenever $\epsilon$ has absolute value less than the size of the smallest nonzero root $\det(A+\epsilon I)\neq 0$.  
