# Doubt regarding groups formation under matrix multiplication

When considering the set of matrices:

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

where

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

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?

• Please don't use bold face all the time. The definition of $Sp(n)$ directly says that $\Omega\subset GL_{2n}(\Bbb R)$, right? Jun 4, 2020 at 13:35
• 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. Jun 4, 2020 at 13:41

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 $$S_1 S_2 \in Sp(n)$$.
• 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 Jun 4, 2020 at 13:54