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Prove that $\det A = 1$ with $A^T M A = M$ and $M = \begin{bmatrix} 0 & I \\ -I &0 \end{bmatrix}$ ($I$ is the identity matrix of order n).

Metin Y.
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KunXian Xia
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    http://math.stackexchange.com/questions/242091/why-is-the-determinant-of-a-symplectic-matrix-1 – Prahlad Vaidyanathan Sep 22 '13 at 11:15
  • could you please post what have you tried till now??? –  Sep 22 '13 at 11:19
  • @Praphulla Koushik I tried to write the matrix $A = \begin{bmatrix} A_1 & A_2 \ A_3 & A_4 \end{bmatrix}$, and get two equalities of those four submatrices. But then I got stuck... – KunXian Xia Sep 22 '13 at 11:47
  • do you know any properties of determinants??? –  Sep 22 '13 at 11:52
  • do you at least see that $det(A)=\pm 1$ ?? (just from the equality $det(AB)=det(A).det(B)$) –  Sep 22 '13 at 11:59
  • this is obvious, I'm stuck because I got no idea on how to exclude the possibility that $\det A = -1$. – KunXian Xia Sep 22 '13 at 12:17
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    $\det A = \pm 1$ ..."this is obvious" ... exactly the sort of thing that should be included in the original question! – GEdgar Sep 22 '13 at 13:12

1 Answers1

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The matrix $A$ that you have described is called a symplectic matrix. The result you are interested in is not trivial, and follows from a series of results. For details, see section 4 of Mackey, Mackey, "On the Determinant of Symplectic Matrices".

The short version is this:

  1. Show that every $A$ is a product of $\mathbb{G}$-reflectors, i.e., matrices of a form $$G = I + \beta u u^T M, \quad \text{for some $\beta \ne 0$, $u \ne 0$}.$$

  2. Show that each $\mathbb{G}$-reflector $G$ has a determinant $\det G = +1$.

Vedran Šego
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