When considering the set of matrices:

$$Sp(n) = \{ S \in \text{GL} (2n, \mathbb{R}) \hspace{2mm} \text{s.t.} \hspace{2mm} S^T \Omega S= \Omega\} \tag{1}$$


$$\tag{2} \Omega = \begin{pmatrix} 0 & \mathbb{I}_{n} \\ - \mathbb{I}_{n} & 0\end{pmatrix}$$

with $\mathbb{I}_{n}$ being the $n \times n$ identity matrix and $0$ denotes the zero matrix of appropriate dimension.

Show that $Sp(n)$ forms a group under matrix multiplication (assuming associativity).

When trying to prove closure I took two matrices into account, labeled $S_1$ and $S_2$.

$$S_1 \in GL(2n, \mathbb{R}) \hspace{5mm} ; \hspace{5mm} S_2 \in GL(2n, \mathbb{R})$$

and therefore $S_1 \cdot S_2 \in GL(2n, \mathbb{R})$

But then I know I must show that these matrices respect the equation above, $S^T \Omega S= \Omega$. How do I do this?

I thought that I needed a representation (general) of an $S$ matrix that had a $det(S) \neq 0 $ but how do I make a general matrix (using letter and not numbers) that clearly shows having a $det \neq 0 $?

Must I also show that $\Omega$ belongs to $GL(2n, \mathbb{R}$? Or is that assumed?

  • 1
    $\begingroup$ Please don't use bold face all the time. The definition of $Sp(n)$ directly says that $\Omega\subset GL_{2n}(\Bbb R)$, right? $\endgroup$ Jun 4, 2020 at 13:35
  • $\begingroup$ I only use it with the purpose of emphasizing the most important parts of the text. Oh, of course it does. Thank you. I have been looking at this question for a long time and didn't notice that. But still, how do I make an example matrix S, do I simply make one that can be written as: $$\begin{pmatrix} a &b \\ c & d \end{pmatrix}$$ and then call $ad-bc \neq 0$ I t just doesn't look like a proof to me. $\endgroup$ Jun 4, 2020 at 13:41

1 Answer 1


It seems your question is: show that $Sp(n)$ forms a group under matrix multiplication.

However, when you want to prove the "closure" you do not have to take $S_1$ and $S_2$ in $GL(2n)$, but in $Sp(n)$.

How to do it: take $S_1 \in Sp(n)$ and $S_2 \in Sp(n)$. Is it true that $S_1 S_2 \in Sp(n)$? Let's try:

$$ (S_1 S_2)^T \Omega (S_1 S_2) = S_2^T S_1^T \Omega S_1 S_2 = S_2^T (S_1^T \Omega S_1) S_2 = S_2^T \Omega S_2 = \Omega $$

This proves that $S_1 S_2 \in Sp(n)$.

  • $\begingroup$ I thought I had to show that a product of two elements also belong in $GL(2n, \mathbb{R})$. It makes perfect sense that way. Thank you $\endgroup$ Jun 4, 2020 at 13:54

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