TITLE: Are positive definite matrices robust to "small changes"?
QUESTION [18 upvotes]: Let $A$ be a positive-definite matrix and let $B$ be some other symmetric matrix. Consider the matrix
$$
C=A+\varepsilon B.
$$
for some $\varepsilon>0$. Is it true that for $\varepsilon$ small enough $C$ is also positive definite?

REPLY [2 votes]: To complement Qiaochu's answer, we have
$$\rm v^\top C \, v = v^\top A \, v + \varepsilon \, v^\top B \, v$$
where $\| \rm v \|_2 = 1$ and $\varepsilon > 0$. Note that
$$\{ \rm v^\top A \, v : \| v \|_2 = 1 \} = [ \lambda_{\min} (\mathrm A), \lambda_{\max} (\mathrm A) ]$$
$$\{ \rm v^\top B \, v : \| v \|_2 = 1 \} = [ \lambda_{\min} (\mathrm B), \lambda_{\max} (\mathrm B) ]$$
and, thus,
$$\{ \rm v^\top C \, v : \| v \|_2 = 1 \} = \left[ \color{blue}{\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B)}, \lambda_{\max} (\mathrm A) + \varepsilon \, \lambda_{\max} (\mathrm B) \right]$$
We know that $\rm A \succ 0$, i.e., $\lambda_{\min} (\mathrm A) > 0$. If $\rm B$ is positive semidefinite, then $\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B) > 0$, i.e., matrix $\rm C$ is positive definite for all values of $\varepsilon > 0$. After all, the conic combination of positive semidefinite matrices is also positive semidefinite. If $\rm B$ is not positive semidefinite, then
$$\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B) = \lambda_{\min} (\mathrm A) - \varepsilon \, |\lambda_{\min} (\mathrm B)| > 0$$
yields the following upper bound on $\varepsilon$
$$\varepsilon < \color{blue}{\frac{\lambda_{\min} (\mathrm A)}{|\lambda_{\min} (\mathrm B)|}}$$