Let V,W be vector space with bases $\{$ $v_1,...,v_n$ $\}$ and $\{$ $w_1,...,w_m$ $\}$ respectively. Let A $=$ $\{$ $T_{kl} : 1\leq k\leq n, 1\leq l \leq m$ $\}$ be a subset of $Hom(V,W)$, where $T_{kl} : V\rightarrow W$ is the unique linear map given as $T_{kl} = w_l$ if $j=k$ and $\textbf{0}$ if $j\neq k$.
I want to show that $A$ is linearly independent. What I did was the following:
Suppose $\sum^{n}_{k=1}$ $\sum^{m}_{l=1}$ $a_{kl}T_{kl} =0$ , where $0$ denotes the zero map. Now, I must show that $a_kl$ =$0$ for each k and each l.
Then $\sum^{n}_{k=1}$ $\sum^{m}_{l=1}$ $a_{kl}T_{kl}(v_j)$ $=$ $\sum^{m}_{l=1}a_{jl}w_j=0$ , as $w_j$ is linearly independent, it follows that $a_{jl}=0$ for each j and each l.
Now, I showed that this holds for each j, and each l how do I show it for each k? What if $j<k$?
You're basically done. Have a think about what your conclusion actually means: you've shown that $a_{jl} = 0$ for each $j$ and each $l$. This is the same as showing $a_{kl} = 0$ for each $k$ and each $l$. You've really just switched up the symbols.