A basic question on linear maps between vector spaces

Question

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 ?

Answer 1

The set you've constructed is linearly independent, but not a basis for $L(V,W)$. Indeed, the mapping $$T_{2,1}:\begin{cases}e_1\mapsto 0\\e_2\mapsto f_1\\e_3\mapsto 0\end{cases}$$ is missing from their span.

What would a full basis for $L(V,W)$ be, then?

Edit: A full basis would be $\{T_{i,j}\mid 1\leq i\leq\dim V, 1\leq j\leq\dim W\}$ where $T_{i,j}(e_i)=f_j$ and for all $m\neq i$ $T_{i,j}(e_m)=0$.