in the book Introduction to symplectic topology from Dusa and McDuff, it claims the following:
Let, $X,Y$ be real $n\times n$ matrices, and let $\Lambda\subset\mathbb{R}^{2n}$ be $\Lambda=\mbox{range }Z$ with Z= $ \begin{pmatrix} X \\ Y \end{pmatrix} $. Then $\Lambda$ is lagrangian iff $X^{T}Y=Y^{T}X$
Then the authors give the following proof: Given $z = (Xu,Yu)$, $z' = (Xu',Yu')$, then $\omega_0=u^{T}(-X^{T}Y+Y^{T}X)u'$. This proves the assertion
However, there are a couple of things I must be missing:
1)This proves that if $X^{T}Y=Y^{T}X$, then $\Lambda\subset\Lambda^*$, but it doesn't prove that there are equal, as claimed
2)The implication to the right, just because $u^{T}(-X^{T}Y+Y^{T}X)u'=0$, this doesn't imply $X^{T}Y=Y^{T}X$. Why is not possibly that $u'$ is just in the kernel of $-X^{T}Y+Y^{T}X$? It doesn't say anywhere that $0$ is not eigenvalue
Hopefully I made myself clear, otherwise feel free to edit it at your will, any help is appreciated.
PS: To clarify definitions, $\mbox{range }Z=\{(Xu,Yu):X\ n\times n \mbox{ matrix and }u\in\mathbb{R}^{2n} \}$
$\omega_0(v,w)=v^TJ_0w$, where $J_0=\begin{pmatrix} 0&-\mathbb{1}\\ \mathbb{1}&0 \end{pmatrix}$ is the symplectic matrix
$\Lambda^*=\{v\in\mathbb{R}^{2n}:\omega_o(v,w)=0,\forall w\in\Lambda \}$
Finally, $\Lambda$ is lagrangian iff $\Lambda=\Lambda^*$
1) You have forgotten the hypothesis that Z has rank n (which is written in the book), hence $\Lambda$'s dimension is n then by equality of dimension you have the equality of vector subspaces (remember that because of the non degeneracy of $\omega_0$, $\mathbb{R}^{2n}$ is the direct sum of $\Lambda$ and $\Lambda^*$)
2) $u^T(-X^TY+Y^TX)u'=0$ for any u,u' thus the bilinear form associated to $-X^TY+Y^TX$ is null so the matrix is zero.