Let $(V,\sigma)$ be a real symplectic vector space and let $\gamma$ be an Euclidean scalar product on $V$. Then there exists a linear isomorphism $A:V\rightarrow V$ such that $$ \gamma(Av,w)=\sigma(v,w) $$ for every $v,w\in V$. It is easy to see that $$ \gamma(Av,w)=-\gamma(v,Aw). $$ This implies that $$ \gamma(A^{2}v,v)=-\gamma(Av,Av)\leq 0. $$ Therefore, if $v\in V, \mu\in\mathbb{R}$ are such that $A^{2}v=\mu v$, then $$ 0\geq \gamma(A^{2}v,v)=\gamma(\mu v,v)=\mu\gamma(v,v), $$ so $\mu\leq 0$. In other words, every eigenvalue of $A^{2}$ is negative. Now, the eigenvalues of $A^{2}$ are precisely the squares of the eigenvalues of $A$. It follows that all the eigenvalues of $A^{2}$ are $0$, which contradicts the fact that $A$ is an isomorphism.
I must be making a mistake somewhere, because they are some steps in the proof of a theorem of certain book, but I do not see where. Can anybody help me?
The eigenvalues of $A$ need not be real ... (in fact they are purely imaginary.)