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Assume $\bf A$ is a (real) positive definite matrix. Let $\varepsilon \neq 0$ be any real (not necessarily positive) number of your choice and $\bf B$ be a fixed (real) symmetric matrix.

Is $\bf A + \varepsilon \bf B$ positive definite? Is it semi-positive-definite? Would anything change if $\bf B$ were not symmetric? Would anything change if $\varepsilon>0$?

The difficulty I'm having is that perhaps $\bf A$ is positive definite but $\bf x^T \bf A \bf x$ can be made arbitrarily small for certain choices of $\bf x$ and so $\bf A + \varepsilon \bf B$ might be negative even though you may choose $\varepsilon$ in advance.

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Assume, as in the question, that $A$ is symmetric (real) positive definite and $B$ is symmetric. All matrices are real. All numbers are real. The set of all matrices on the form $A+\epsilon B$ is called a matrix pencil, there is a wikipedia aricle on that (too short). This can be used to define a generalized eigenvalue problem: $$ Bv=\lambda A v $$ solutions $\lambda $ are called generalized eigenvalues and solutions $v$ are called generalized eigenvectors (note that this terms also has other meanings!). We can define a generalized Rayleigh ratio as $$ R(B,A,x) = \frac{x^T Bx}{x^T Ax} $$ Since A is positive definite, this will always be defined when $x \not=0$. Let $\delta$ be the minimum generalized eigenvalue. Then we have, in analogy with the situation for the ordinary Rayleigh quotient (there is a Wikipedia article), that $$ R(B,A,x) \ge \delta \text{~~when $x\not= 0$.} $$ What does $\delta$ say about $B$? First, note that $\delta=0$ then $B$ is positive semidefinite (and singular). In that case we can see that $$ x^T(A+\epsilon B)x >0 $$ as long as $\epsilon \ge 0$. Then we can look at the case $\delta >0$: Let $\mu_0>0$ be the smallest eigenvalue of $A$. Vi har $$ x^T Bx \ge \delta x^T Ax \ge \delta \mu_0 x^Tx $$ slik at $\begin{multline} \frac{x^T (A+\epsilon B) x}{x^T x} = (x^T Ax+\epsilon x^T Bx )/x^Tx \\ \ge (x^T Ax +\epsilon \delta \mu_0 x^Tx)/x^T x \\ \ge (\mu_0 x^Tx + \epsilon \delta \mu_0 x^T x)/x^T x \\ \ge \mu_0 (1+\epsilon \delta) > 0 \end{multline} $ and solving the inequality for $\epsilon$ gives $$ \epsilon > -\frac{1}{\delta} $$ so we can conclude that $A+\epsilon B$ is positive definite if that inequality is fulfilled.

You can solve for the case $\delta <0$ in like manner!

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  • $\begingroup$ Doesn't solving for $\epsilon$ give $\epsilon > -1/\delta$? $\endgroup$ – Pierre-Guy Plamondon Feb 3 '14 at 17:19
  • $\begingroup$ Thanks kjetil b halvorsen and Pierre-Guy Plamondon for your comments and response. I wrote up another solution (see below) based on the link math.stackexchange.com/questions/226486/…. $\endgroup$ – user103828 Feb 4 '14 at 8:30
  • $\begingroup$ Pierre-Guy Plamondon: Corrected now! $\endgroup$ – kjetil b halvorsen Feb 4 '14 at 15:52
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Let $\bf A$ be a positive definite real matrix. All matrices, vectors, and numbers are real. Let $\bf x$ be a vector satisfying $|| \bf x||=1$. Then by positive definiteness of $\bf A$ and compactness of the set where $|| \bf x||=1$, there is some $\delta>0$ such that $$ {\bf x^T A x} > \delta \qquad (*) $$ Now choose $\varepsilon$ so that $\varepsilon ({\bf x^T B x}) \geq -\delta$. Then $$ {\bf x^T (A + \varepsilon B ) x} = {\bf x^T A x} + \varepsilon {\bf x^T B x} > \delta - \delta=0 $$ This leads to $A + \varepsilon \bf B$ being positive definite on the unit ball and hence positive definite (and hence also semi positive-definite).

Furthermore, nothing changes if $\bf B$ were just an arbitrary finite square matrix or $\varepsilon$ is restricted to be positive (or negative). Nevertheless, the proof fails if $\bf A$ is semi positive-definite (and not positive-definite) because $(*)$ does not hold.

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  • $\begingroup$ This is not correct. start with checking your strange equation $\delta - 1/\delta =0$! $\endgroup$ – kjetil b halvorsen Feb 4 '14 at 15:59
  • $\begingroup$ oops. i corrected the equation. $\endgroup$ – user103828 Feb 4 '14 at 16:57

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