What's a basis for $\mathcal L(V,W)$?

I understand that the dimension of $\mathcal L(V,W)$ is dim$(V)$dim$(W)$ because there is an isomorphism with $\mathbb F^{m,n}$ where $m=$ dim$(V)$ and $n=$ dim$(W)$. This means that there is a basis for $\mathcal L(V,W)$ with dim$(V)$dim$(W)$ elements, right? What's a possible basis?

• See the answers to this question Jan 24, 2018 at 15:47
• @Omnomnomnom Oh, so the $L_{ij}(v_k)=\delta_{ik}w_j$ are a basis for it. Thank you. Jan 24, 2018 at 16:02

Once you've chosen a basis $(u_i)_{1\le i\le\dim V}$ for $V$ and a basis $(v_j)_{1\le j\le\dim W}$ for $W$, a basis for $\mathcal L(V,W)$ is made up of the linear maps $\bigl(\varepsilon_{ij}\bigr)_{\substack{1\le i\le\dim V,\\[0.5ex] 1\le j\le\dim W}}$ defined by $$\begin{cases}\varepsilon_{ij}(u_i)=v_j,\\[0.5ex] \varepsilon_{ij}(u_k)=0&\text{if }k\ne i.\end{cases}$$

Yes. And the vector basis are matrices $E_{i,j}$ such that the $i-j$ th entry is 1 and the others 0:

$$\begin{pmatrix} 0 & 0 & \cdots & 0 &\cdots & 0\\ 0 & 0 & \cdots & 0 &\cdots & 0 \\ \vdots & \vdots &\ddots & \vdots &\ddots & \vdots\\ 0 & 0 & \cdots & 1 & \cdots 0\\ \vdots & \vdots &\ddots & \vdots &\ddots & \vdots\\ 0 & 0 & \cdots & 0 &\cdots & 0\\ \end{pmatrix}$$

EDIT

$\mathcal L(V,W)$ is not actually the space if matrices $n\times m$, so they cannot be a basis. My answer should be read in the sense that, a basis for $\mathcal L(V,W)$ are the operators $\varphi_{i,j}$ such that, given basis $\{v_i\},\{w_j\}$ of $V$ and $W$ respectively, their matrix representation are the $E_{i,j}$.

• The space in question is a space of linear transformations, not a space of matrices. They are isomorphic, fine, but this doesn't quite answer the question. Jan 24, 2018 at 15:54
• @DavidHill: Ok. So think in the operators $\varphi_{i,j}$ such that in the basis $\{v_i\}$ of $V$ and $\{w_j\}$ of $W$ have the above matrix representation. Jan 24, 2018 at 16:02