# How to find a base for the set of multilinear functions?

Let $$L(V_1,V_2,...,V_n;\mathbb{R})$$ the set of multilinear functions where $$V_i$$ are finite vector fields. Find a basis for $$L(V_1,V_2,...,V_n;\mathbb{R})$$.

I've proved that $$L(V_1,V_2,...,V_n;\mathbb{R})$$ is a vector field and for any $$G, F \in L(V_1,V_2,...,V_n;\mathbb{R}) \implies G\otimes F\in L(V_1,V_2,...,V_n;\mathbb{R})$$

But i don't know how to find a basis for that vector space, how i asociate the basis of each $$V_i$$ with the basis of $$L(V_1,V_2,...,V_n;\mathbb{R})$$?

• Is an element of $L(V_1 \times \cdots \times V_n; \mathbb{R})$ a function of $V_1 \times \cdots \times V_n$ to itself or to $\mathbb{R}$? – Connor Harris Sep 30 '19 at 17:49
• @ConnorHarris to $\mathbb{R}$, question fixed. – sango Sep 30 '19 at 17:50
• en.wikipedia.org/wiki/Dual_basis – Connor Harris Sep 30 '19 at 17:54

I just consider two $$\mathbb{R}$$-linear spaces $$V$$ and $$W$$ in order to simplify the notation, but this works in general.
Let $$\{e_1,...,e_n \}$$ be a basis of $$V$$ and let $$\{f_1,...,f_m\}$$ be a basis of $$W$$.
For any $$i \in \{1,...,n \}$$ and any $$j \in \{1,...,m \}$$, let $$F_{i,j}$$ be the element of $$L(V,W;\mathbb{R})$$ such that $$F_{i,j}(v,w)=v_iw_j$$ for every $$(v,w) \in V\times W$$, being $$v_i$$ the $$i$$-th component of $$v$$ (with respect to the given basis) and being $$w_j$$ the $$j$$-th component of $$w$$ (with respect to the given basis).
Then $$\{F_{i,j}: i \in \{1,...,n \}, j \in \{1,...,m \}\}$$ is a basis of $$L(V,W;\mathbb{R})$$.
For the linear independence, observe that, if $$\lambda_{1,1},\lambda_{1,2},...\lambda_{n,m}$$ are real numbers such that $$\sum_{i,j}\lambda_{i,j} F_{i,j}$$ is the null element of $$L(V,W,\mathbb{R})$$, then $$0=\sum_{i,j}\lambda_{i,j}F_{i,j}(e_{i'},f_{j'})=\lambda_{i',j'}$$ for every $$i' \in \{1,...,n\}$$ and every $$j' \in \{1,...,m\}$$.
For the generation, observe that, for every $$F\in L(V,W,\mathbb{R})$$, it is the case that: \begin{aligned} F(v,w)&=F(v_1e_1+...+v_ne_n,w_1f_1+...+w_mf_m) \\ &=F(e_1,f_1)v_1w_1+...+F(e_n,f_m)v_nw_m\\ &=F(e_1,f_1)F_{1,1}(v,w)+...+F(e_n,f_m)F_{n,m}(v,w) \end{aligned} for every $$(v,w)\in V\times W$$. Hence $$F=F(e_1,f_1)F_{1,1}+...+F(e_n,f_m)F_{n,m}$$.