symplectic lie algebra is simple The symplectic lie algebra defined by $sp\left(n\right)=\left\{ X\in gl_{2n}\,|\, X^{t}J+JX=0\right\}$  when $J=\begin{pmatrix}0 &amp; I\\
-I &amp; 0\end{pmatrix}$. So $X\in sp\left(n\right)$ is of the sort $X=\begin{pmatrix}A &amp; B\\
C &amp; -A^{t}\end{pmatrix}$ when $B=B^{t},\, C=C^{t}$.

This far I was able to get, but how can I prove that it is simple? 

Thanks!
Let ${\mathfrak g}=sp(n)$.

Short answer, from an advanced viewpoint : $\mathfrak g$ is semisimple (because $K(x,y)=tr(xy)$ defines a nondegenerate invariant bilinear form on $\mathfrak g$), so it will be simple iff its Dynkin diagram is connected. But we know that the diagram is $C_n$ and this is clearly connected.

Longer answer, from a more pedestrian viewpoint : a standard basis $\cal B$ for $\mathfrak g$ is $\lbrace h_i,a_i,b_i\rbrace_{1 \leq i \leq n} \cup \lbrace c_{i,j},d_{i,j},f_{i,j},g_{i,j}\rbrace_{1 \leq i&lt;j \leq n}$ where

$$
h_i=e_{ii}-e_{n+i,n+i}, \ \ \ a_i=e_{i,i+n}, \ \ \  b_i=e_{i+n,i},
$$

$$
c_{i,j}=e_{ij}-e_{i+n,j+n}, \ \ \ d_{i,j}=e_{ji}-e_{j+n,i+n},
$$

$$
f_{i,j}=e_{i,j+n}-e_{j,i+n}, \ \ \  g_{i,j}=e_{i+n,j}-e_{j+n,i} 
$$

(and $e_{ij}$ is the matrix all of whose coefficients are zero except for the $(i,j)$-th which is $1$). 

Now let $\mathfrak i$ be an ideal of $\mathfrak g$. Then $\mathfrak i$ is stable by $ad(x)$ for any $x\in {\mathfrak g}$. In particular, if $H$ is the vector space spanned by 
$h_1,h_2, \ldots ,h_n$, then $\mathfrak i$ is stable by all the $ad(h)$, for
$h \in H$. Now those $ad(h)$ are simultaneously diagonalized by the basis we have just described above. We deduce that $\mathfrak i=H&#39; \oplus {\sf Vect}(F)$, where $H&#39;$ is a linear subspace of $H$ and $F$ is a subset of ${\cal B}&#39;={\cal B} \setminus \lbrace h_1,h_2, \ldots ,h_n \rbrace$. Let us now consider the root spaces
$$
AB_i={\sf Vect}(h_i,a_i,b_i), \ CD_{ij}={\sf Vect}(h_i-h_j,c_{ij},d_{ij}), \
FG_{ij}={\sf Vect}(h_i+h_j,f_{ij},g_{ij})
$$
Each root space $R$ is a Lie subalgebra of $\mathfrak g$ isomorphic to ${\mathfrak sl}_2$. It easily follows that either ${\mathfrak i} \cap R=\lbrace 0 \rbrace$ or 
$R \subseteq {\mathfrak i}$ for any $R$. Combining this with the decomposition of $\mathfrak i$  above, we see that $\mathfrak i$ is the direct sum of some root spaces, plus some subspace  of $H$. 

Suppose $F$ contains $a_i$ for some $i$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i,a_i,b_i)$. For $j&gt;i$, $\mathfrak i$ also contains 
$[a_i,g_{ij}]=c_{ij}$, $h_i-[c_{ij},d_{ij}]=h_j, [h_j,\frac{a_j}{2}]=a_j$. Similarly, for $k&lt;i$,
$\mathfrak i$ also contains $[a_i,-g_{ki}]=d_{ki}$, $h_i+[c_{ki},d_{ki}]=h_k, [h_k,\frac{a_k}{2}]=a_k$. So $\mathfrak i$ contains all the $a_i$, and we easily deduce that $\mathfrak i=A$. 

Suppose $F$ contains $c_{ij}$ for some $i&lt;j$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i-h_j,c_{ij},d_{ij})$. And $\mathfrak i$ also contains $h_i+h_j=[c_{ij},f_{ij}]$. So $\mathfrak i$ contains both $h_i$ and $h_j$, it also contains the root space
${\sf Vect}(h_i,a_i,b_i)$, so $\mathfrak i=A$ by the preceding paragraph.

Suppose $F$ contains $f_{ij}$ for some $i&lt;j$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i+h_j,f_{ij},g_{ij})$. And $\mathfrak i$ also contains $c_{ij}=[f_{ij},b_{j}]$, so $\mathfrak i=A$ by the preceding paragraph.

Finally we see that $\mathfrak i=A$ unless $F$ is empty. In that case, $\mathfrak i$ is a linear subspace of $H$. For any $i$, $ad (a_i)$ must be zero on $\mathfrak i$, otherwise 
$\mathfrak i$ would contain $a_i$. It is easily deduced that ${\mathfrak i}=\lbrace 0 \rbrace$,
so $\mathfrak g$ is indeed simple.
$sp(n)$ is a Lie subalgebra of the more familiar lie algebra $sl(n)$ because $Sp(n)$ is a Lie subgroup of $SL(n)$ (determinant $= 1$ condition). Since $sl(n)$ is semisimple, and therefore has non-degenerate Killing form, $tr(xy)$ would also be non-degenerate on its lie subalgebras.