= Solution
For a <symplectic vector space> $(V,\omega)$ and a <vector subspace> $W$, the <symplectic orthogonal complement> is
$$
W^\omega=\{v\in V:\omega(v,w)=0\text{ for every }w\in W\}.
$$
<Nondegenerate bilinear form> identifies $V$ with its <dual space>, so $\dim W^\omega=\dim V-\dim W$. If $W$ is a <symplectic subspace>, $W\cap W^\omega=0$, and hence $V=W\oplus W^\omega$; the restriction to $W^\omega$ is also nondegenerate.
Here is an inductive construction of a <symplectic basis>. Choose $e_1\ne0$. <Nondegenerate bilinear form> supplies $f_1$ with $\omega(e_1,f_1)=1$. The plane $P=\operatorname{span}(e_1,f_1)$ is a <symplectic subspace>, and its <symplectic orthogonal complement> gives $V=P\oplus P^\omega$. Repeat on $P^\omega$. The process ends in dimension zero, since each step removes two dimensions and a one-dimensional alternating form is degenerate. We obtain $e_1,f_1,\ldots,e_n,f_n$ with
$$
\omega(e_i,e_j)=\omega(f_i,f_j)=0,\qquad\omega(e_i,f_j)=\delta_{ij}.
$$
The <linear map> taking the coordinate vectors of $\mathbb R^{2n}$ to this basis pulls $\omega$ back to $\sum_jdx_j\wedge dy_j$, proving the asserted normal form.
Now split $V=C\oplus D$ with $D=C^\omega$, and choose <symplectic bases> on both planes. Parametrize $\Gamma_A$ by $v\mapsto(Av,v)$. Orthogonality removes the cross terms, and a two-dimensional alternating form transforms by the <determinant>, so
$$
\omega|_{\Gamma_A}\longleftrightarrow A^*\omega_C+\omega_D=(1+\det A)\omega_D.
$$
Consequently the correct <nondegenerate bilinear form> criterion is
$$
\boxed{\Gamma_A\text{ is symplectic}\iff\det A\ne-1.}
$$
The strict inequality printed in the PDF requires an extra orientation condition: it says that the <symplectic orientation> on the graph agrees with the orientation transported from $D$. It is not equivalent to being a <symplectic subspace>. For example, $A=\operatorname{diag}(-2,1)$ has <determinant> $-2$ and gives a nondegenerate restriction $-\omega_D$. This is the <orientation criterion for a graph of symplectic planes>.
This same map supplies the requested obstruction. Orient $C$ and $\Gamma_A$ by their restricted <symplectic forms>. If $(d_1,d_2)$ is a positive basis of $D$, then $(Ad_1+d_1,Ad_2+d_2)$ is negatively oriented in $\Gamma_A$. The coordinate transformation from a basis of $C$ followed by these graph vectors to a basis of $C\oplus D$ has <determinant> $1$. Reversing the graph basis to give its positive orientation therefore makes the oriented direct sum $C\oplus\Gamma_A$ negative relative to the ambient <symplectic orientation>.
Two distinct complex lines in standard $\mathbb C^2$ have the opposite behaviour: complex bases of the two lines give a complex-linear <isomorphism> $\mathbb C\oplus\mathbb C\to\mathbb C^2$, whose real <determinant> is the positive number $|\det_{\mathbb C}|^2$. Their complex orientations are their <symplectic orientations>. A <symplectic linear map> preserves both the ambient and each plane's <symplectic orientation>, so it cannot change this sign. Thus \b[the pair with $A=\operatorname{diag}(-2,1)$ cannot be the image of a pair of complex lines]. Interchanging the planes does not change the sign, since both dimensions are even.
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