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Suppose $V_1$ and $V_2$ are $k$-vector spaces with bases $(e_i)_{i \in I}$ and $(f_j)_{j \in J}$, respectively. I've seen the claim that the collection of elements of the form $e_i \otimes f_j$ (with $\left(i,j\right) \in I \times J$) forms a basis for $V_1 \otimes V_2$. But I seem to get stuck with the proof.

My question: What's the easiest way to see that the above set is indeed linearly independent?

darij grinberg
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JPSR
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3 Answers3

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Construct a set $\{\phi_{i,j}\}$ of linear forms on the tensor product which is a dual basis to the family $\{e_i \otimes f_j\}$ (in the sense that for each pair $\left(i, j\right) \in I \times J$, the map $\phi_{i,j}$ sends $e_i \otimes f_j$ to $1$ while sending all $e_{i'} \otimes f_{j'}$ with $\left(i', j'\right) \neq \left(i, j\right)$ to $0$). That will immediately imply linear independence.

If $\{\psi_i\}$ is a dual basis to your basis $\{e_i\}$ of $V_1$ and $\{\rho_j\}$ is a dual basis to your basis $\{f_j\}$ of $V_2$, then you can consider the map $\phi_{i,j} = \psi_i\otimes \rho_j:V_1\otimes V_2\to k\otimes k\cong k$ for each pair $\left(i, j\right) \in I \times J$. This map $\phi_{i,j}$ sends $e_i \otimes f_j$ to $1$ while sending all $e_{i'} \otimes f_{j'}$ with $\left(i', j'\right) \neq \left(i, j\right)$ to $0$.

darij grinberg
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Mariano Suárez-Álvarez
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    I've added a few more details to this answer. Note that "dual basis" is a bit of a misleading notion here, since it will not be a basis of the dual space when $V_1 \otimes V_2$ is infinite-dimensional. – darij grinberg Feb 03 '19 at 05:28
  • I find this answer misleading, as it gives the wrong impression that a map can be defined on the tensor product directly, without further justifications (e.g. using the universal property). – MCL Sep 23 '22 at 10:10
  • This answer defines maps on a tensor product simply by tensoring them. I did not check that that works, just as I did not check that the tensor product exists... – Mariano Suárez-Álvarez Sep 23 '22 at 22:20
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First, take linear functions $\delta_i$ defined on $V_1$ and $\varphi_i$ defined on $V_2$ such that $$\delta_i(e_i)=1$$ $$\delta_i(e_j)=0 \ \ i\neq j$$ $$\varphi_i(f_i)=1$$ $$\varphi_i(f_j)=0 \ \ i\neq j.$$ Now, consider the bilinear form $\phi_{ij}:V_1\times V_2\to \mathbb{R}$ given by $\phi_{ij}(v_1,v_2) = \delta_i(v_1)\cdot\varphi_j(v_2)$. The universal property gives a linear function $\widetilde{\phi_{ij}}:V_1\otimes V_2\to \mathbb{R}$ that maps $e_i\otimes f_j$ to $1$ and any other $e_{i'}\otimes f_{j'}$ with $i\neq i'$ or $j\neq j'$ to $0$, then they are linearly independent.

Teodora
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I'll only show that it spans $V_1\otimes V_2$.

Consider $U=\operatorname{span}\{e_i\otimes f_j:i\in I,j\in J\}$, with $\{e_i:i\in I\}$ basis for $V_1$ and $\{f_j:j\in J\}$ basis for $V_2$.

We have trivially that $U$ is a subspace of $V_1\otimes V_2$.

Let $h:V\times W \to V\otimes W$ be the bilinear map associated to the tensor product via its universal property. That is, $h(v,w)=v\otimes w$.

It is clear that $\operatorname{im}(h)=\operatorname{span}\{h(e_i,f_j)=e_i\otimes f_j: i\in I,j\in J\}=U$. But, in this case, $(U,h)$ also satisfies the universal property of tensor product. Thus, $U\cong V_1\otimes V_2$. Then, $U=V_1\otimes V_2$.

Thus, $V_1\otimes V_2$ is generated by $\{e_i\otimes f_j:i\in I,j\in J\}$.

Arctic Char
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Gomes93
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