Basis for the tensor product

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Important for manifolds or something

Statement

Let $\mathbb{F}$ be a field and let $\big((V_i,\mathbb{F})\big)_{i\eq 1}^k$ be a family of finite dimensional vector spaces. Let $n_i:\eq\text{Dim}(V_i)$ and $e^{(i)}_1,\ldots,e^{(i)}_{n_i}$ denote a basis for $V_i$, then we claim:

• $$\mathcal{B}:\eq\left\{e^{(1)}_{i_1}\otimes\cdots\otimes e^{(k)}_{i_k}\ \big\vert\ \forall j\in\{1,\ldots,k\}\subset\mathbb{N}[1\le i_j\le n_j]\right\}$$

Is a basis for the tensor product of the family of vector spaces, $V_1\otimes\cdots\otimes V_k$

Note that the number of elements of $\mathcal{B}$, denoted $\vert\mathcal{B}\vert$, is $\prod_{i\eq 1}^kn_i$ or $\prod_{i\eq 1}^k\text{Dim}(V_i)$, thus:

• $\text{Dim}(V_1\otimes\cdots\otimes V_k)\eq\prod_{i\eq 1}^k n_i$

Proof

1. The proposed "basis" actually spans $V_1\otimes\cdots\otimes V_k$, i.e $V_1\otimes\cdots\otimes V_k\subseteq\text{Span}(\mathcal{B})$
   • Let $A\in V_1\otimes\cdots\otimes V_k$ be given. Then:
     • there is an $m\in\mathbb{N}$ such that $$A\eq\sum_{\alpha\eq 1}^m \lambda_\alpha (v_{\alpha,1}\otimes\cdots\otimes v_{\alpha,k})$$ for some $\lambda_\alpha\in\mathbb{F}$ and $v_{\alpha,i}\in V_i$
     • But each $$v_{\alpha,j}\eq\sum^{n_j}_{i_j\eq 1}v_{\alpha,j,i_j}e^{(j)}_{i_j}$$ (where $e^{(j)}_{i_j}$ is the $i_j$^th basis vector of $V_j$ and $v_{\alpha,j,i_j}$ the $i_j$^th coefficient of $v_{\alpha,j}$)
     • Thus: $$A\eq\sum^m_{\alpha\eq 1}\lambda_\alpha\left(\left(\sum_{i_1\eq 1}^{n_1}v_{\alpha,1,i_1}e^{(1)}_{i_1}\right)\otimes\cdots\otimes\left(\sum_{i_k\eq 1}^{n_k}v_{\alpha,k,i_k}e^{(k)}_{i_k}\right)\right)$$
     • $$\eq\sum^m_{\alpha\eq 1}\lambda_\alpha\sum^{n_1}_{i_1\eq 1}v_{\alpha,1,i_1}\left(e^{(1)}_{i_1}\otimes\left(\sum_{i_2\eq 1}^{n_2}v_{\alpha,2,i_2}e^{(2)}_{i_2}\right)\otimes\cdots\otimes\left(\sum_{i_k\eq 1}^{n_k}v_{\alpha,k,i_k}e^{(k)}_{i_k}\right)\right)$$
     • $$\eq\sum^m_{\alpha\eq 1}\sum^{n_1}_{i_1\eq 1}\lambda_\alpha v_{\alpha,1,i_1}\left(e^{(1)}_{i_1}\otimes\left(\sum_{i_2\eq 1}^{n_2}v_{\alpha,2,i_2}e^{(2)}_{i_2}\right)\otimes\cdots\otimes\left(\sum_{i_k\eq 1}^{n_k}v_{\alpha,k,i_k}e^{(k)}_{i_k}\right)\right)$$
     • $$\eq\sum^m_{\alpha\eq 1}\ \underbrace{\sum^{n_1}_{i_1\eq 1}\cdots\sum^{n_k}_{i_k\eq 1} }\ \lambda_\alpha\ \underbrace{ v_{\alpha,1,i_1}\cdots v_{\alpha,k,i_k} }\ \left(e^{(1)}_{i_1}\otimes\cdots\otimes e^{(k)}_{i_k}\right)$$ where the elements with the underbrace range over $\{1,\ldots,k\}$
     • $$\eq\sum^m_{\alpha\eq 1}\ \underbrace{\sum_{\begin{array}{c}i_1,\ldots,i_k \\ \forall j\in\{1,\ldots,k\}[1\le i_j\le \text{Dim}(V_j)]\end{array} } }_{\text{Finitely many - }\left(\prod^k_{\gamma\eq 1}\text{Dim}(V_\gamma)\right)\text{ - terms} }\ \overbrace{\lambda_\alpha\ \prod_{\beta\eq 1}^k v_{\alpha,\beta,i_\beta} }^{\in\mathbb{F} }\ \underbrace{\left(e^{(1)}_{i_1}\otimes\cdots\otimes e^{(k)}_{i_k}\right)}_{\in\mathcal{B} }$$
   • So it is clear that $V_1\otimes\cdots\otimes V_k\subseteq \text{Span}(\mathcal{B})$
2. Linear independence

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At least do outline, I've done it once

References

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Smooth Manifolds, and Linear Algebra via Exterior Products