i think the question is actually this: if $\displaystyle [x_i,x_j]=\sum_{k=1}^n c_{ij}^k x_k,$ and

$\displaystyle X_i = \sum_{r,s} c_{ri}^sa_{rs}, \ \ X_j=\sum_{u,v}c_{uj}^v a_{uv},$

then

$\displaystyle [X_i,X_j]=\sum_{k=1}^n c_{ij}^k X_k. \ \ \ \ \ \ \ \ (1)$

for all $\displaystyle i,j.$ to prove this, first recall that the scalars $\displaystyle c_{ij}^k$ are called the **structure constants** of the Lie algebra $\displaystyle L$ and they satisfy the following relations:

(i) $\displaystyle c_{ij}^k = -c_{ji}^k,$ for all $\displaystyle 1 \leq i,j,k \leq n.$

(ii) $\displaystyle \sum_{s=1}^nc_{is}^qc_{jp}^s + c_{js}^qc_{pi}^s + c_{ps}^qc_{ij}^s = 0$, for all $\displaystyle 1 \leq i,j,p,q \leq n.$

now $\displaystyle [a_{rs},a_{uv}]=\delta_{su}a_{rv}-\delta_{rv}a_{us}$ with (i) and (ii) will give us:

$\displaystyle [X_i,X_j]=\sum_{r,s,u,v} c_{ri}^sc_{uj}^v(\delta_{su}a_{rv}-\delta_{rv}a_{us}) = \sum_{r,s,v}c_{ri}^sc_{sj}^va_{rv}-\sum_{r,s,u}c_{ri}^sc_{uj}^ra_{us}=\sum_{p,q,s} (c_{sj}^qc_{pi}^s - c_{si}^qc_{pj}^s)a_{pq}=\sum_{p,q} \left (\sum_{s=1}^n c_{pi}^sc_{sj}^q-c_{si}^qc_{pj}^s \right)a_{pq}=\sum_{p,q,s}c_{ps}^qc_{ij}^sa_{pq}=$

$\displaystyle \sum_{s=1}^n c_{ij}^s \sum_{p,q} c_{ps}^qa_{pq}=\sum_{s=1}^n c_{ij}^sX_s. \ \Box$