Are positive definite matrices robust to "small changes"?

Question

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?

Answer 1

Positive definite means that $\langle v, Av \rangle > 0$ for all nonzero vectors $v$; actually it suffices to check this condition for unit vectors. We have

$$\langle v, Cv \rangle = \langle v, Av \rangle + \epsilon \langle v, Bv \rangle.$$

Now, by the compactness of the unit sphere, $\langle v, Av \rangle$ takes on a minimum nonzero value $m$ on unit vectors (the smallest eigenvalue of $A$, although we don't need this), and $| \langle v, Bv \rangle |$ takes on a maximum nonzero value $M$ on unit vectors (the largest eigenvalue of $B$ in absolute value).

So we can take $\epsilon < \frac{m}{M}$, which gives

$$\langle v, Cv \rangle \ge m - \epsilon M > 0$$

for all unit vectors $v$. So $C$ is positive definite as desired.

Answer 2

If by "positive definite" you mean "strictly positive definite", the answer is "yes". The set of strictly positive definite matrices is an open set in the space of symmetric matrices.

For the following reason. The positive definite matrices are the ones which satisfy a certain finite set of determinental inequalites (the principal minor determinants must all be strictly positive), each one of of which cuts out an open set in the space of matrices. Alternatively, from first principles, let $X$ be the closed unit sphere in vector space, let $Y$ be the symmetric matrices. The function $(v,A)\mapsto \|Av\|$ is continuous, so the set $S=\{(v,A): \|Av\|\le0\}\subset X\times Y$ is closed. Since $X$ is compact, the map $\pi:(v,A)\mapsto A$ is a closed map, so $\pi(S)$ is closed in $Y$, so the complement of $\pi(S)$ is open in $Y$. But that complement is the set of all $A$ for which $\|Av\|>0$ for all $v\in X$, that is to say, the set of all (strictly) positive definite matrices.

Answer 3

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 $\varepsilon$

$$\varepsilon < \color{blue}{\frac{\lambda_{\min} (\mathrm A)}{|\lambda_{\min} (\mathrm B)|}}$$