Let $gl(n)$ be spanned by $n^2$ abstract operators $\{a_j^i\}$ satisfying the $gl(n)$ commutation relation:
$$[a_j^i,a_l^k] = a^i_l \delta_j^k - a_j^k \delta_l^i$$
Why can we conclude that the subspace of diagonal operators $\{a_k^k\}$ forms a maximal commutative subalgebra?
It's the maximality I am asking about, I can see that $[a^j_j,a^k_k]=0,\forall k,j$
I guess lets see if we can put a non-diagonal operator into this, so $$[a_j^i,a_l^k] = a^i_l \delta_j^k - a_j^k \delta_l^i,\quad i\ne j,k \ne l$$
We can get $k\ne j,i\ne l$ and it commutes, so $[a_1^2,a_3^4]= a^2_3\delta^4_1-a_1^4\delta_3^2=0$
So yes, why is it the maximal commutative subalgebra
This is maximal because to add another element would require that element to commute with all of the diagonal generators, which we can see that for any $a_j^k$ there is a diagonal operator $a_k^k$(or $a_j^j$ would do) such that we get \begin{align} [a_j^k,a_k^k]=&0\\ \implies& a^k_k\delta^k_j-a_j^k\delta_k^k=0\\ \implies& a^k_k\delta^k_j=a^k_j\delta_k^k=a^k_j \end{align} So we obtain that either $a_j^k=0$ or $a_j^k=a^k_k$. In the first case we get a contradiction, since this is one of the $n^2$ generators, and in the latter case we see that it is a diagonal operator.