I wrote this in my notes:

Let $V_1,\ldots,V_k$ be real vector spaces of dimensions $n_1,\ldots,n_k$ and let $V_j$ have basis $(\mathbf{e}^{(j)}_1,\ldots,\mathbf{e}^{(j)}_{n_j})$ and corresponding dual basis $(\mathbf{e}_{(j)}^1,\ldots,\mathbf{e}_{(j)}^{n_j})$. Then the set $$ \mathcal{B} = \{\mathbf{e}_{(1)}^{i_1}\otimes\cdots\otimes\mathbf{e}_{(k)}^{i_k}\colon 1\leq i_1\leq n_1,\ldots,1\leq i_k\leq n_k\} $$ is a basis for $L(V_1,\ldots,V_k;\mathbb{R})$ with dimension $n_1\cdots n_k$. For each ordered $k$-tuple $(i_1,\ldots,i_k)$ of integers with $1\leq i_j \leq n_j$ define the numbers $$ F_{i_1,\ldots,i_k} = F(\mathbf{e}_{i_1}^{(1)},\ldots,\mathbf{e}_{i_k}^{(k)}) $$ then the basis expansion for $F\in L(V_1,\ldots,V_k;\mathbb{R})$ is $$ F = F_{i_1,\ldots,i_k}\mathbf{e}_{(1)}^{i_1}\otimes\cdots\otimes\mathbf{e}_{(k)}^{i_k} $$ using the summation convention.

And realized It said $L(V_1,\ldots,V_k;\mathbb{R})$ and not $L(V_1,\ldots,V_k;W)$, with $W$ a vector space of finite dimension.

My question is, what do I have to do to use the above formula, if my multilinear mapping is vector valued and not scalar valued? Perhaps the formula "just works", but the coefficients are vectors in $W$ instead of scalars? Or perhaps one need to somehow make the target space $W$ into one of the tensor products? Would like constructive answers with respect to bases. Thank you.


Note that for $F \in L(V_1, \dots, V_k; W)$, we have $F_{i_1,\ldots,i_k} = F(\mathbf{e}_{i_1}^{(1)},\ldots,\mathbf{e}_{i_k}^{(k)}) \in W$. Let $\mathbf{w}_1, \dots, \mathbf{w}_m$ be a basis for $W$, then $F_{i_1,\dots, i_k} = F_{i_1,\dots, i_k}^l\mathbf{w}_l$ where $F_{i_1,\dots, i_k}^l \in \mathbb{R}$ for $l = 1, \dots, m$. So now

$$F = F_{i_1,\ldots,i_k}^l\mathbf{e}_{(1)}^{i_1}\otimes\cdots\otimes\mathbf{e}_{(k)}^{i_k}\otimes\mathbf{w}_l.$$

  • $\begingroup$ OK, I will go with that! $\endgroup$ – Emil Jul 11 '18 at 17:09

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