= Solution
The <Poisson bracket> in the original coordinates is
$$
\boxed{\{F,G\}_{q,p}=\sum_{j=1}^n(F_{q_j}G_{p_j}-F_{p_j}G_{q_j}).}
$$
For $z=(q,p)$, set
$$
J=\begin{pmatrix}0&I_n\\-I_n&0\end{pmatrix},\qquad A=Df(z).
$$
Then $\omega_0(u,v)=u^T Jv$ and $\{F,G\}=\nabla F^T J\nabla G$. The pullback definition of a <canonical transformation> gives the pointwise <matrix> condition
$$
\boxed{A^TJA=J.}
$$
Since $A$ is invertible, this is equivalent to $AJA^T=J$. Indeed, invert $A^TJA=J$ and use $J^{-1}=-J$ to obtain $A^{-1}JA^{-T}=J$, then multiply by $A$ and $A^T$. Applying the same calculation to the inverse implication gives equivalence. These are the two equivalent <symplectic matrix> identities that connect the <symplectic form> with <Poisson brackets>.
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