Prove that $\mathcal{B}$ is a basis for $\mathcal{L}(V , W)$

Question

Let $V$ and $W$ two vector spaces on a field $K$ and let ${\{{\phi}_i\}}_{i = 1}^n$ and ${\{w_j\}}_{j = 1}^m$ be two basis for $V^*$ and $W$ respectively. Defining, for each $(\phi , w) \in V^* \times W$, $$ \begin{array}{llccl} {\Psi}_{(\phi , w)} & : & V & \to & W \\ & & v & \mapsto & \phi(v) w\mbox{,} \end{array} $$ is $\mathcal{B} = {\{{\Psi}_{({\phi}_i , w_j)}\}}_{(i , j) \in \{1 , \ldots , n\} \times \{1 , \ldots , m\}}$ a basis for the vector space $\mathcal{L}(V , W)$ formed by the linear maps between $V$ y $W$? Thank you very much in advance.

Answer 1

Take $T\colon V \to W$ linear. Assume $(v_1,\ldots,v_n)$ is the basis of $V$ dual to $(\phi_1,\ldots,\phi_n)$. For each $v \in V$, we have $v= \sum_{j=1}^n \phi_j(v)v_i$, by definition of dual basis. Then $$T(v) = \sum_{j=1}^n \phi_j(v) T(v_j) = \sum_{i=1}^m \sum_{j=1}^n \phi_j(v)a_{ij}w_i,$$where $[T]_{(v_i),(w_j)} = (a_{ij})$ is the matrix of $T$ with respect to these bases. Well, this writes as $$T(v) = \sum_{i=1}^m \sum_{j=1}^n a_{ij} \Psi_{(\phi_j ,w_i)}(v),$$so $$T = \sum_{i=1}^m \sum_{j=1}^n a_{ij} \Psi_{(\phi_j,w_i)}$$and we conclude that $(\Psi_{(\phi_i,w_j)})_{(i,j) \in \{1,\ldots,n\}\times \{1,\ldots,m\}}$ is a spanning set for ${\rm Lin}(V,W)$. If you already know that $\dim {\rm Lin}(V,W) = nm$, we're done.

Otherwise, you have to check for linear independence directly. One does this as follows: Assume that $$\sum_{i=1}^m \sum_{j=1}^n a_{ij} \Psi_{(\phi_j,w_i)} = 0.$$We must check that $a_{ij} = 0$ for all possibilities of $i$ and $j$. Evaluate the above expression in $v_k$: $$0 =\sum_{i=1}^m \sum_{j=1}^n a_{ij} \phi_j(v_k)w_i = \sum_{i=1}^m \sum_{j=1}^n a_{ij}\delta_{jk}w_i = \sum_{i=1}^m a_{ik}w_i,$$and linearly independence of $(w_i)_{i=1}^m$ gives $a_{ik} = 0$. But $k$ was also arbitrary, so we're done.