In optical ray tracing it's possible to use symplectic matrices. I have a problem with them. If a matrix $M$ is symplectic, this means that for $M$ the following equation hols: $$M^T\Omega M=\Omega$$ where $$\Omega = \begin{pmatrix} 0 & I_n & \\ -I_n & 0 \end{pmatrix}$$ The determinant of $M$ is one: $$det(M)=1$$ If I have a product of symplectic matrices: $$M_t=\prod_{k=1}^NM_k$$ the determinant of the product is the same way $1$. So, how is it possible to prove: $$det(M_t)=1$$ Thanks
2 Answers
Use $det(AB...Z)=det(A)det(B)...det(Z)$ and the fact that $det(M_k)=1$.
I am not sure what exactly you are asking. If you ask why $\det \prod_{k=1}^N M_k=1$ when $\det M_1,\ldots,\det M_N=1$, this is because the determinant function is multiplicative (i.e. $\det AB=(\det A)(\det B)$). If you are asking why a sympletic matrix has determinant $1$, this has been explained in the related Wikipedia article. If you are asking why $\prod_{k=1}^N M_k$ is sympletic, this is because \begin{align*} (M_1\cdots M_k)^T\Omega(M_1\cdots M_k) &=(M_k^T\cdots M_1^T)\Omega(M_1\cdots M_k) \\ &=M_k^T\left(M_{k-1}\left(\cdots\left(M_2^T(M_1^T\Omega M_1)M_2\right)\cdots \right)M_{k-1}\right)M_k\\ &=\Omega. \end{align*}