Dimension of Lie algebra

Let $o(2l,F)$, with $l \ge2$ and $n=2l$ be the orthogonal lie algebra $\{L\in gl(n,F)|SL=-L^{t}S\}$ where $S=\begin{pmatrix} 0 &I_{l} \\ I_{l} & 0 \end{pmatrix}$. How can I show that $\dim(o(2l,F))=2l^{2}-l$?

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How are $l$ and $n$ linked? – Davide Giraudo Jan 31 '12 at 18:45
It doesn't seem to make sense unless $n=2l$. I'm more curious about the "$s$" that shows up unintroduced in the last formula. – Henning Makholm Jan 31 '12 at 18:59
If instead we consider $\{L\in F^{2l\times 2l}\mid SL=-L^t S\}$, then we do get a Lie algebra with the standard bracket $[A,B]=AB-BA$. – Henning Makholm Jan 31 '12 at 19:12
@HenningMakholm, sorry for the confusion, I have edited the question. – Edison Jan 31 '12 at 19:32

We write $L=\pmatrix{A&B\\ C&D}$. Then $L\in o(2l,F)$ iff $Sl=-L^tS$. We have $$SL=\pmatrix{0&I\\ I&0}\pmatrix{A&B\\ C&D}=\pmatrix{C&D\\ A&B}$$ and $$-L^tS=\pmatrix{A^t&C^t\\ B^t&D^t}\pmatrix{0&I\\ I&0}=\pmatrix{-C^t&-A^t\\ -D^t&-B^t}$$ so $C$ and $B$ have to be skew-symmetric and $D=-A^t$. So $L=\pmatrix{A&0\\ 0&-A^t}+\pmatrix{0&B\\ 0&0}+\pmatrix{0&0\\ C&0}$. Denoting $E_{rc}$ the matrix whose entries are $0$ except the term of the row $r$ and column $c$. Then $o(2l,F)$ is generated by $$\left\{\pmatrix{E_{rc}&0\\\ 0&E_{cr}},1\leq c,r\leq l\right\}\cup\left\{\pmatrix{0&E_{rc}-E_{cr}\\\ 0&0},1\leq c<r\leq l\right\}\cup\left\{\pmatrix{0&0\\\ E_{rc}-E_{cr}&0},1\leq c<r\leq l\right\},$$ which is linearly independent.
The dimension of $l\times l$ skew-symmetric matrices is $\frac{l(l-1)}2$ so $$\dim o(2l,F)=2\frac{l(l-1)}2+l^2=l^2-l+l^2=2l^2-l.$$