The dual of a Fréchet space.

Let $\mathcal{F}$ be a Fréchet space (locally convex, Hausdorff, metrizable, with a family of seminorms $\{\|~\|_n\}$).

I've read that the dual $\mathcal{F}^*$ is never a Fréchet space, unless $\mathcal{F}$ is actually a Banach space. I'd like to know in which ways the dual can fail to be Fréchet in general; for example, is it always incomplete as a metric space? Or is it always non-metrizable? What are the real issues here?

If there's a relatively easy proof of this fact (that $\mathcal{F}^*$ is not Fréchet unless $\mathcal{F}$ is Banach), I would appreciate it as well.

Thanks.

[EDIT: The reference cited by Dirk contains a Theorem whose proof is inaccessible to me, so I'd upvote/accept an answer that at least sketches such a proof, or provides another way of settling the question.

I'd also be interested in any further explanations regarding the statement that "(...) a LCTVS cannot be a (non-trivial) projective limit and an inductive limit of countably infinite families of Banach spaces at the same time.", made by Andrew Stacey in that link, which is related to my question.]

The answer depends a bit on what you mean by "is a Frechet space": If you consider $F^*$ just as a vector space there might be complete norms on it which, however, have nothing to do with the duality. If $F$ itself is not normed there is no complete norm $p$ on $F^*$ whose topology is finer than the weak-star-topology: Since every continuous linear functional on $F$ is continuous with respect to some $\|\cdot\|_n$ one has $F^*=\bigcup_n A_n$ where $A_n=\{f\in F^*: |F| \le n\|\cdot\|_n \}$ is weak-star-compact by Alaoglu's theorem and hence $p$-closed. Now, Baire's theorem implies that some $A_N$ has $p$-interior points and the convexity of $A_N$ implies that $0$ is an interior point. This implies $F^* = \bigcup_m m A_N$, that is, every $f\in F^*$ is continuous with respect to the same seminorm $\| \cdot\|_N$. This only holds for normed spaces.