Consider the set of all linear maps from $V$ to $W$ with dimension $n$ and $m$ respectively. This forms a vector space under the addition and scalar multiplication defined by $(T + S)(v) = T(v) + S(v)$ and $(\alpha T)(v)=\alpha T(v)$. Now it is said that the dimension of this vector space is $mn$.
Now for $n=3, m=2$ I see that the following two maps $T_1$ and $T_2$ satisfy the basis constraints :
$$ T_1(e_1) = f_1, T_1(e_2) = f_1, T_1(e_3) = f_2 $$ $$ T_2(e_1) = f_2, T_2(e_2) = f_2, T_2(e_3) = f_1 $$
To show that any linear map can be written as a linear combination of this two maps, Let T be any such linear map.
Now, $$T(e_1)=\alpha_1 f_1 + \alpha_2 f_2 =\alpha_1 T_1(e_1) + \alpha_2 T_2(e_1) =(\alpha_1 T_1)(e_1) + (\alpha_2 T_2)(e_1) =(\alpha_1 T_1 + \alpha_2 T_2)(e_1)$$
Similar thing holds for $T(e_2)$ and $T(e_3)$ also.
And, regarding the linearly independence let $$(\alpha T_1 + \beta T_2)(e_1) = 0 => \alpha T_1(e_1) + \beta T_2(e_1) =0 => \alpha f_1 + \beta f_2 =0 => \alpha=0 \beta=0 $$
where $e_i$s and $f_i$s are basis vectors of $V$ and $W$ respectively. Where am I going wrong ?