I read following conclusion from a paper:
Say $X$ is a positive definite $n \times n$ matrix, if $\|Y\|_2 < \sigma_n(X)$, where $\sigma_n(X)$ is the smallest eigenvalue of $X$, then $X+Y$ is also positive definite.
Why is that?
both $X$ and $Y$ are symmetric matrix here.
A general matrix $A$ is said to be positive definite if and only if $$A + A^T$$ is symmetric positive definite or equivalently if and only if $$ u^T(A+A^T) u > 0, \quad u \not = 0.$$
If $A$ is symmetric, then this condition is obviously equivalent to $$u^T A u > 0, \quad u \not = 0.$$ As pointed out by David C. Ullrich, the equivalence hold regardless of whether $A$ is symmetric or not, because $$u^TAu = (u^TAu)^T = u^T A^T u,$$ which implies $$ u^T (A + A^T) u = 2 u^T A u. $$
In our case, both $X$ and $Y$ are symmetric and our goal is to show that $Z = X + Y$ is symmetric positive definite, i.e. $$ u^TZ u > 0, \quad u \not = 0.$$ By the spectral theorem, we have $$ u^T X u \ge \lambda_{\min}(X) \|u\|^2,$$ where $\lambda_{\min}(X) > 0$ denotes the smallest eigenvalue of $X$. Cauchy-Schwartz's inequality implies that $$ |u^T Y u| \leq \|Y\| \|u\|^2.$$ In particular, we have $$ u^T Y u \ge - |u^T Y u| \ge - \|Y\| \|u\|^2,$$ from which it follows that $$ u^TZu = u^T X u + u^T Y u \ge \lambda_{\min}(X)\|u\|^2 - \|Y\| \|u\|^2 = (\lambda_{\min}(X) - \|Y\|) \|u\|^2. $$ By assumption, $\|Y\| < \lambda_{\min}(X)$, so $$ u^TZu > 0, \quad u \not = 0.$$ It follows, that $Z$ is symmetric positive definite.
In popular terms, if the perturbation $Y$ is small enough, then $Z = X + Y$ is certainly nonsingular and the threshold is determined by the smallest eigenvalue of $X$.