Phase space volume doesn't change under canonical transform

Question

I have given a set of generalized coordinates $(q_1,..q_n,p_1,..p_n)$. Suppose I had a canonical transform $(q_i,p_i)\rightarrow (Q_i,P_i).$ I am trying to show that the phase space volume element remains constant after the transform. I'm trying to understand the proof given in https://courses.smp.uq.edu.au/MATH4104/m4104sec3.pdf (Property 2)

I don't understand how $AJ_nA^T=\begin{bmatrix} \{Q_i,Q_j\} & \{Q_i,P_j\} \\ \{P_i,Q_j\} & \{P_i,P_j\} \\ \end{bmatrix}.$

For example for the first block matrix/first quadrant $(1\leq i,j\leq n)$ I get $$(AJ_nA^T)_{ij}=\left(-\sum_{k=1}^n\frac{\partial Q_i}{\partial p_k}\right)\left(\sum_{k=1}^n\frac{\partial Q_j}{\partial q_k}\right)+\left(\sum_{k=1}^n\frac{\partial Q_i}{\partial q_k}\right)\left(\sum_{k=1}^n\frac{\partial Q_j}{\partial p_k}\right)= $$ $$ =\{Q_i,Q_j\}+x. $$ $x$ only has terms $\partial Q_j/\partial p_i$ and $\partial Q_j/\partial q_i$ where $i\neq j$.

Answer 1

Multiply by blocks:

$ A= \left[ {\begin{array}{cc} \partial _{q}Q & \partial _{p}Q \\ \partial _{q}P & \partial _{p}P \\ \end{array} } \right] $, $ A^{t}= \left[ {\begin{array}{cc} (\partial _{q}Q)^{t} & (\partial _{p}Q)^{t} \\ (\partial _{q}P)^{t} & (\partial _{p}P)^{t} \\ \end{array} } \right]$ and therefore $JA^{t}= \left[ {\begin{array}{cc} (\partial _{p}Q)^{t} & (\partial _{p}P)^{t} \\ -(\partial _{q}Q)^{t} & -(\partial _{q}P)^{t} \\ \end{array} } \right]$.

Now for example $(AJA^{t})_{11}=\partial _{q}Q(\partial _{p}Q)^{t}-\partial _{p}Q(\partial _{q}Q)^{t}$.

Remeber this is an equation on matrices. Write the matrices explicitly, for example $(\partial _{q}Q)_{ij}=\frac {\partial Q_{i}} {\partial q_{j}} $,$(\partial _{p}Q)^{t}_{ij}=\frac {\partial Q_{j}} {\partial q_{i}}$, and put them into the equation above, perform the matrix multiplication and get the Poisson brackets.