Show $L(V,W)$ is generated by a some elementary linear mappings

Let $\mathbb {K}$ be a field, $V, W$ two $\mathbb {K}-$ vector spaces. Let $v_1,...v_m$ be a basis in $V $ and $w_1, w_2,...w_n$ a basis in $W.$ We define the mappings $$F_{i,j}: V \rightarrow W,\,\, F_{i,j}(\sum_{k=1}^m \lambda_k v_k) = \lambda_i w_j, i=1,...m, j=1,...n.$$ I need to show that $F_{i,j}$ is a basis of $L(V,W).$

We know that dim$L(V,W) = mn.$ Thus we have the right number of $F_{i,j}$ linear mappings. We have to show that $F_{i,j}$ is generating $L(V,W)$ and that $F_{i,j}$ are linearly independent, i.e. $\sum_{i=1}^m \lambda_i F_{i,j} + \sum_{j=1}^n \lambda_j F_{i,j} = 0 \implies \lambda_i = \lambda_j = 0 \,\,\,\forall i =1,...m, j= 1,...n.$

Let $f \in L(V,W).$ We know that given the two basis in $V$ and $W$, $f$ is fully determined, i.e. $f(v_i) = \sum_{j=1}^n a_{ij}w_j, i=1,...m.$ Thus, $f(\sum_{k=1}^m \lambda_k v_k) = \sum_{k=1}^m \lambda_k f(v_k) = \sum_{k=1}^m \sum_{j=1}^n \lambda_k a_{kj} w_j.$ I do not see how to include here the functions $F_{i,j}.$

Can somebody provide some proposal how to go further or provide a solution proposal ? Thanks.

Solution 1:

Hint: Note that two linear maps $f,g$ over $V$ are equal if and only if $f(v_i) = g(v_i)$ for $i=1,\dots,n$. Given coefficients $a_{ij}$ such that $f(v_{i}) = \sum_{j=1}^n a_{ij}w_j$, look for coefficients $b_{p,q}$ such that the function $g = \sum_{p,q} b_{p,q}F_{p,q}$ satisfies $$ \left[\sum_{p,q} b_{p,q}F_{p,q}\right](v_i) = f(v_i) $$ for $i = 1,\dots,m$. In so doing, you have shown that the set $\{F_{ij}\}$ spans $L(V,W)$.

To show that the functions are linearly independent, consider a linear combination $g = \sum_{p,q} b_{p,q}F_{p,q}$. Note that $g = 0$ holds if and only if $g(v_i) = 0$ for $i = 1,\dots,n$.

Solution 2:

Another approach can also be through this $L(V,W)\cong M_{n\times m}(\mathbb{K})$.

For the given bases $\mathcal{B,B'}$ of $V$ and $W$. The isomorphism is given by $\psi:L(V,W)\to M_{n\times m}(\mathbb{K}) $ . such that $\psi(T)=[T]_{\mathcal{B,B'}}$ ( The matrix of $T$ wrt to these given basis).

Where $T$ is an arbitrary element of $L(V,W)$.

and the basis for $M_{n\times m}(\mathbb{K})$ is given by $\{E_{ij}\}_{1\leq i\leq m,1\leq j\leq n}$ . Where $E_{ij}$ is the matrix such that the $ij$th entry is $1$ and all others are $0$. Its very easy to prove that these matrices form a basis for $M_{n\times m}$ and infact it is the standard basis for this space. Then the maps corresponding to these matrices would form a basis for $L(V,W)$.

Verify that your $F_{ij}$'s correspond to these $E_{ij}$'s. IT is easy to do this. we have $F_{i,j}(v_{i})=w_{j}$ . So the $i$-th column of $[F_{i,j}]_{\mathcal{B,B'}}$ is say $(0,0,...1,..,0)^{T}$ where the $1$ occurs in the $j$-th position. And $F_{i,j}(v_{k})=0$ for $k\neq i$. So the Matrix of this map $F_{i,j}$ is just a matrix whose $ij$-th entry is $1$ and the rest are $0$.

Then it would immediately follow that your $F_{i,j}$'s form a basis for $L(V,W)$ as an isomorphism maps a basis to a basis.

$F_{i,j}=\psi^{-1}(E_{ij})$ and as $\psi^{-1}$ is also an isomorphism we have $\{F_{i,j}\}_{1\leq i\leq n,1\leq j\leq m}=\{\psi^{-1}(E_{ij})\}_{1\leq i \leq n\,1\leq j\leq m}$ is a basis for $L(V,W)$.

$$\mathbb{Q.E.D}$$