If $B(q)$ is a symmetric matrix and $B(0)$ is positive definite, is $T = p^T B(q) p$ positive definite?

Question

In book "Stability Theory by Lyapunov's Direct Method" by Laloy there is a General Hypotheses concerning stability, which says:

"The kinetic energy is $T(q, p) = \frac{1}{2} p^TB(q)p$, where $B(q)$ is symmetric and $C^1$, and $B(0)$ is positive definite. [...]"

If $B(0)$ is positive definite, then $p^TB(0)p$ is bigger than or equal zero for any p, but does it also means that $p^TB(q)p$ is bigger than or equal zero for any p?

If $B(0)$ is positive definite, is $B(q)$ positive definite?

I appreciate any help.

Answer 1

Unless we have more information about $B(q)$, the answer is no. For instance, consider $$ B(q) = \pmatrix{1 & 0\\0 & 1 - q} $$ and consider the vector $p = (0,1)^T$.