subspaces of a symplectic vector spaces are of special forms.

Let $$(V,\omega)$$ be a symplectic vector space. Let $$F \subseteq V$$ be a subspace.

Show that $$V$$ admits a symplectic basis $$\{e_1,\ldots,e_n,f_1,\ldots,f_n\}$$ with the following properties:

(1) If $$F$$ is symplectic then $$F=span\{e_1,\ldots,e_k,f_1,\ldots,f_k\}$$ for some $$k$$

(2) If $$F$$ is isotropic then $$F=span\{e_1,\ldots,e_k\}$$ for some $$k$$

(3) If $$F$$ is co-isotropic then $$F=span\{e_1,\ldots,e_n,f_1,\ldots,f_k\}$$ for some $$k$$

(4) If $$F$$ is Lagrangian then $$F=span\{e_1,\ldots,e_n\}$$

The converse of these statements are easy to prove by applying definitions. But I have no idea how to prove these.

I'm guessing this question is unanswered because it's an exercise and you haven't sketched any work on the problem. The ideas for all 4 are basically the same, and all ideas you need to do it is the linear algebra "tinkering" that you do in the (symplectic) Gram-Schmidt process. I would recommend reading a proof of the symplectic Gram-Schmidt theorem, then trying to prove these four statements using the same ideas. You can probably find a lot of places to read about symplectic GS, or I just wrote an explanation in my answer to $$SP_{2n}(\mathbb {R})$$ acts transitively on $$\mathbb {R}^{2n}$$. If you get stuck, you can update the question with your progress and we'll try to help you out.