[proofplan] The proof is a dimension count using the symplectic orthogonal complement. Isotropicity gives the inclusion $W\subset W^\omega$, and the [dimension formula for symplectic complements](/theorems/10039) then forces $\dim W\le n$. Coisotropicity is handled by applying the same isotropic bound to $W^\omega$, while the Lagrangian case follows by substituting $W=W^\omega$ into the same dimension formula. The converse uses isotropicity to get $W\subset W^\omega$ and equality of dimensions to upgrade this inclusion to equality. [/proofplan]

[step:Use isotropicity and the symplectic complement formula to bound $\dim W$ from above]Assume that $W$ is isotropic. By definition of isotropic subspace, $W\subset W^\omega$. Hence \begin{align*} \dim W\le \dim W^\omega. \end{align*} By [citetheorem:10039], applied to the subspace $W\subset V$, \begin{align*} \dim W+\dim W^\omega=\dim V=2n. \end{align*} Combining the two displayed relations gives \begin{align*} 2\dim W\le \dim W+\dim W^\omega=2n. \end{align*} Therefore $\dim W\le n$.[/step]

[guided]Assume that $W$ is isotropic. The definition of isotropic subspace says precisely that every vector in $W$ is symplectically orthogonal to every vector in $W$. In terms of the symplectic orthogonal complement, this is the inclusion \begin{align*} W\subset W^\omega. \end{align*} Since $V$ is finite-dimensional, inclusion of subspaces implies the corresponding dimension inequality: \begin{align*} \dim W\le \dim W^\omega. \end{align*} Now we use the structural input about symplectic complements. The theorem [citetheorem:10039] applies because $W$ is a subspace of the finite-dimensional symplectic [vector space](/page/Vector%20Space) $(V,\omega)$. It gives \begin{align*} \dim W+\dim W^\omega=\dim V. \end{align*} The hypothesis of the theorem states that $\dim V=2n$, so \begin{align*} \dim W+\dim W^\omega=2n. \end{align*} The point of comparing $W$ with $W^\omega$ is that the dimension formula now turns the inclusion into a numerical bound. Since $\dim W\le \dim W^\omega$, we have \begin{align*} 2\dim W\le \dim W+\dim W^\omega. \end{align*} Substituting the dimension formula gives \begin{align*} 2\dim W\le 2n. \end{align*} Dividing by $2$ yields \begin{align*} \dim W\le n. \end{align*}[/guided]

[step:Apply the isotropic bound to the symplectic complement of a coisotropic subspace] Assume that $W$ is coisotropic. By definition, $W^\omega\subset W$. We first check that $W^\omega$ is isotropic. Let $u,v\in W^\omega$. Since $u\in W^\omega\subset W$ and $v\in W^\omega$, the definition of $W^\omega$ gives \begin{align*} \omega(v,u)=0. \end{align*} Because $\omega$ is alternating, \begin{align*} \omega(u,v)=-\omega(v,u)=0. \end{align*} Thus $W^\omega$ is isotropic. Applying the isotropic bound already proved to the subspace $W^\omega\subset V$ gives \begin{align*} \dim W^\omega\le n. \end{align*} Using [citetheorem:10039] again, \begin{align*} \dim W+\dim W^\omega=2n. \end{align*} Therefore \begin{align*} \dim W=2n-\dim W^\omega\ge 2n-n=n. \end{align*} So $\dim W\ge n$. [/step]

[step:Substitute the Lagrangian equality into the dimension formula] Assume that $W$ is Lagrangian, so by definition \begin{align*} W=W^\omega. \end{align*} Applying [citetheorem:10039] to $W\subset V$ gives \begin{align*} \dim W+\dim W^\omega=2n. \end{align*} Substituting $W^\omega=W$ gives \begin{align*} 2\dim W=2n. \end{align*} Hence $\dim W=n$. [/step]

[step:Upgrade a maximal-dimensional isotropic subspace to a Lagrangian subspace] Conversely, assume that $W$ is isotropic and that $\dim W=n$. Isotropicity gives \begin{align*} W\subset W^\omega. \end{align*} By [citetheorem:10039], \begin{align*} \dim W+\dim W^\omega=2n. \end{align*} Since $\dim W=n$, this becomes \begin{align*} n+\dim W^\omega=2n, \end{align*} so $\dim W^\omega=n$. Thus $W\subset W^\omega$ and the two finite-dimensional subspaces have the same dimension. Therefore \begin{align*} W=W^\omega. \end{align*} By definition, $W$ is Lagrangian. This proves the converse and completes the proof. [/step]