View source for Basis for the tensor product

{{Stub page|grade=A*|msg=Important for manifolds or something}}
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==[[/Statement|Statement]]==
{{/Statement}}
==Proof==
# The proposed "basis" actually spans {{M|V_1\otimes\cdots\otimes V_k}}, i.e {{M|V_1\otimes\cdots\otimes V_k\subseteq\text{Span}(\mathcal{B}) }}
#*Let {{M|A\in V_1\otimes\cdots\otimes V_k}} be given. Then:
#** there is an {{M|m\in\mathbb{N} }} such that {{MM|A\eq\sum_{\alpha\eq 1}^m \lambda_\alpha (v_{\alpha,1}\otimes\cdots\otimes v_{\alpha,k})}} for some {{M|\lambda_\alpha\in\mathbb{F} }} and {{M|v_{\alpha,i}\in V_i}}
#** But each {{MM|v_{\alpha,j}\eq\sum^{n_j}_{i_j\eq 1}v_{\alpha,j,i_j}e^{(j)}_{i_j} }} (where  {{M|e^{(j)}_{i_j} }} is the {{M|i_j}}<sup>th</sup> basis vector of {{M|V_j}} and {{M|v_{\alpha,j,i_j} }} the {{m|i_j}}<sup>th</sup> coefficient of {{M|v_{\alpha,j} }})
#** Thus: {{MM|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)}}
#** {{MM|\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)}}
#** {{MM|\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) }}
#** {{MM|\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 {{M|\{1,\ldots,k\} }}
#** {{MM|\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 {{M|V_1\otimes\cdots\otimes V_k\subseteq \text{Span}(\mathcal{B})}}
# [[Linearly independent set|Linear independence]]
{{Requires proof|grade=A*|msg=At least do outline, I've done it once}}
==References==
<references/>
{{Requires references|grade=A|msg=Smooth Manifolds, and Linear Algebra via Exterior Products}}
{{Theorem Of|Linear Algebra|Abstract Algebra}}