I want to show that the lie algebra $\mathfrak{sp}(2n)$ is semisimple by showing that its Killing form is non degenerate. This is a very direct method and presumably there are simpler ones, but I want to make do with what I've got. $\mathfrak{sp}(2n)$ has the basis;
$$\begin{align*} \mathcal{B}= &\{H_{i}:=E_{i,i}-E_{n+i,n+i}\mid 1\leq i\leq n\}&\cup\\ &\{E_{i,n+j}+E_{j,n+i},E_{n+i,j}+E_{n+j,i}\mid 1\leq i<j\leq n\}&\cup\\ &\{E_{i,j}-E_{n+j,n+i}\mid 1\leq i\neq j\leq n\}&\cup\\ &\{E_{i,n+i},E_{n+i,i}\mid 1\leq i\leq n\} \end{align*}$$
The first set is a Cartan Subalgebra of $\mathfrak{sp}(2n)$, and the other sets consist of distinct root vectors of $\mathfrak{sp}(2n)$. Using some elementary theorems about roots and computing the $ad(H_{i})$ eigenvalues of the basis elements I can see that in this basis the cartan subalgebra will correspond to an $n\times n$ block of the Killing form matrix, and the other terms will correspond (with the correct ordering of the basis), to $2\times 2$ blocks. So if we can show these blocks are invertible we will be done.
It is fairly easy to get an exression for the Killing form on the Cartan subalgebra and thus show the $n\times n$ block is invertible, but I'm struggling to show that this also holds for the $2\times 2$ blocks.
For example $\{E_{i,n+i},E_{n+i,i}\}$ should correspond to a $2\times 2$ block for each $i=1,...,n$ as can be seen from the $ad(H_{i})$ eigenvalues. I have that;
$$ \kappa(E_{i,n+i},E_{n+i,i})=tr(ad(E_{i,n+i})ad(E_{n+i,i})) $$
I have computed $ad(E_{i,n+i})$ in this basis but even for $n=2$ this is a $10\times 10$ matrix, and since the basis isnt indexed nicely I have been unable to multiply it to compute $\kappa$.
If I could somehow find the $ad(E_{i,n+i})ad(E_{n+i,i})$ eigenvalues, that would also work, but that idea hasn't gotten me very far.
If I can figure out how to show each $2\times 2$ block is invertible I'll be done. However I just can't seem to get to the bottom of this computation. Any help would be much appreciated.
We can compute the trace of $\mathrm{ad}_{E_{i,n+i}} \mathrm{ad}_{E_{n+i,i}}$ in the basis you gave. For in the standard basis $\{E_{ij}\}$ of matrices, we can compute $$ \mathrm{ad}_{E_{ij}}(E_{kl}) = \delta_{jk} E_{il} - \delta_{li} E_{kj},$$ where $\delta_{ij}$ is the Kronecker delta.
For example, if $i,j\leq n$, $$\mathrm{ad}_{E_{i,n+i}} \mathrm{ad}_{E_{n+i,i}} (E_{j,j} - E_{n+j,n+j}) = \mathrm{ad}_{E_{i,n+i}} \left( E_{n+i,i}\delta_{ij} + E_{n+i,i} \delta_{ij}\right) = 2\delta_{ij} \left( E_{i,i} - E_{n+i,n+i}\right).$$ Hence, the contribution to the trace $\kappa(E_{i,n+i},E_{n+i,i})$ from the first part of your basis is $2$.
Hopefully this gives enough insight to proceed.