TITLE: Does $(\mathbf A+\epsilon \mathbf I)^{-1}$ always exist? Why?
QUESTION [13 upvotes]: 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?

REPLY [6 votes]: If $A$ is symmetric positive semidefinite, and $\epsilon > 0$, then $A + \epsilon I$ is symmetric positive definite (and hence invertible).
To see this, note that if $x \neq 0$ then
\begin{align}
x^T(A + \epsilon I) x &= x^T A x + \epsilon x^T x \\
&= x^T A x + \epsilon \|x\|^2 \\
& > 0.
\end{align}