Hmmm... I think that it's perfectly appropriate to say that V(x)=5 has a minimum; Its minimum is "5" (at every

x). It just doesn't have a (unique) minimum

point, which is a stronger condition then merely having a minimum (value).

However, in the context of your solution, I'll grant you that "

V has a minimum", although technically ambiguous, is most naturally interpreted as "

V has a minimum point at equilibrium".

However, the actual issue here is that

V can have a minimum point at equilibrium, while

might nonetheless

not be positive definite - as you already demonstrated with your

example. So the "if and only if" part of your solution's final sentence is incorrect. (As is my first suggested correction to it, in my second post from the top of the thread).

A corrected version of it would be:

V has a minimum point at equilibrium if is positive definite;

is non-negative definite if V has a minimum point at equilibrium.---------

About metastability you are correct - I apologise. I'm afraid I had the term "metastable" confused with "neutrally stable" (or "critically stable"/"marginally stable"), which may

sometimes be used to describe situations such as the example I gave, for linear systems. (It does entirely depend on the definition of "stability", though, like you said).

Also this would generally

only apply to the

linearised system, not the actual system, because (like you already pointed out) the actual stability of the system when

is non-negative definite with (at least one) zero eigenvalue, cannot be determined from the linearisation alone but must be determined by taking higher order terms of the Taylor expansion into account.