# What is a basis for the space of multilinear maps from $V_1 \times \dots \times V_k \to W$?

I know that a basis for the space $$L(V_1, \dots, V_k; \mathbb R)$$ is $$\{\varepsilon^{i_1} \otimes \cdots \otimes \varepsilon^{i_k}\mid 1 \le \varepsilon^{i_j} \le \dim(V_j)\}$$ where $$\varepsilon^{i_j}$$ is the dual basis vector to the basis vector $$E_{i_j}$$ for $$V_j$$.

But what if we change this to $$L(V_1, \dots, V_k; W)$$. What would be a basis for this space?

• If $W$ is $n$-dimensional, then $W\cong \mathbb{R}^n$; maps into $\mathbb{R}^n$ are in one-to-one correspondence with families of $n$ maps into $\mathbb{R}$; so you get a direct product of $n$ copies of the space $L(V_1,\ldots,_k;\mathbb{R})$; and then you can use the general facts about how to get a basis for $X\times Y$ when you have a basis for $X$ and a basis for $Y$. – Arturo Magidin Nov 17 '18 at 0:09
• @ArturoMagidin So, $L(V_1, \dots, V_k; \mathbb R^n) \cong \prod_{i=1}^k L(V_1, \dots, V_k; \mathbb R)$? – Al Jebr Nov 17 '18 at 1:00
• Yes; this is true because of the universal property of the product. Careful if $W$ is infinite dimensinoal, though, because then it is isomorphic to the direct sum of copies of $\mathbb{R}$, and maps into the direct sum are not in one-to-one correspondence with families of maps. – Arturo Magidin Nov 17 '18 at 1:07
• A natural basis to take is $\{\epsilon^{i_1} \otimes \cdots \otimes \epsilon^{i_k} \otimes w_j\}$ where the $\epsilon^j$ range over a basis for the dual of $V_j$, and the $w$ ranges over a basis for $W$. It's a very natural extension of what you already have: each $\epsilon$ eats a vector from a $V$, and rather than getting a number out the other side, you get a vector of $W$. – Joppy Nov 17 '18 at 1:24
• @Joppy But how does $\{\epsilon^{i_1} \otimes \cdots \otimes \epsilon^{i_k} \otimes w_j\}$ act on vectors? The element $\epsilon^{i_1} \otimes \cdots \otimes \epsilon^{i_k} \otimes w_j$ takes in $k$ vectors, but there are $k+1$ positions in that tensor product. – Al Jebr Nov 18 '18 at 17:45