Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $V_1$ and $V_2$ are $k$-vector spaces with bases $(e_{i1})$ and $(e_{i2})$, respectively. I've seen the claim that the collection of elements of the form $e_{i1} \otimes e_{i2}$ 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?

share|cite|improve this question
They should be $e_{i1}\otimes e_{j2}$. – user18119 Nov 3 '11 at 13:50

Construct a set $\{\phi_{i,j}\}$ of linear forms on the tensor product which is a dual basis. That immediaely implies linear independence.

If $\{\psi_i\}$ is a dual basis to your basis of $V_1$ and $\{\rho_j\}$ is a dual basis to your basis of $V_2$, then you can consider the map $\phi_{i,j}=\psi_i\otimes \rho_j:V_1\otimes V_2\to k\otimes k\cong k$.

share|cite|improve this answer

Just making what Mariano Suárez-Alvarez wrote explicit:

Let $V_1$ be a vector space with basis $\{e_i\}$ and $V_2$ a vector space with basis $\{f_j\}$. It's easy to see that $\{(e_i,f_j)\}$ is a basis for $V_1\times V_2$. Since $V_1 \otimes V_2$ is a quotient of $V_1 \times V_2$, the projection of the basis elements of $V_1 \times V_2$ must span $V_1 \otimes V_2$. (Edit: this is false! Please see Mariano and Alex's comments below.)

Let $\phi_{i,j}: V_1 \times V_2 \rightarrow k \;$ be the dual basis for $V_1 \times V_2$. That is, $\phi_{i,j}((e_i,f_j)) = 1$ and $f_{i,j} = 0$ on any other basis element. Note that each $\phi_{i,j}$ is bilinear.

Since $F$ is also an $F$-vector space, the universal property of tensor products implies that for each $\phi_{i,j}$ there is a unique homomorphism $\Phi_{i,j}: V_1 \otimes V_2 \rightarrow F$ such that $\Phi_{i,j}(e_i\otimes f_j) = \phi_{i,j} ((e_i,f_j))$.

Suppose $\sum_{i,j} c_{i,j} e_i \otimes f_j = 0$ for some scalers $c_{i,j}$ not all equal to 0. Then $\Phi_{k,l}(\sum_{i,j} c_{i,j} e_i \otimes f_j ) = c_{k,l}$ is not equal to 0 for some $(k,l)$. But $\Phi_{k,l}(\sum_{i,j} c_{i,j} e_i \otimes f_j ) = \Phi_{k,l}(0) = 0$. So no such $c_{i,j}$ exist, and the set of $e_i\otimes f_j$ is linearly independent.

share|cite|improve this answer
Can you be a little more explicit in your argument where you conclude that the basis of $V_1 \times V_2$ spans the tensor product of these spaces? Does this follow from some property of quotient maps in general? – ItsNotObvious Nov 3 '11 at 14:25
$V_1\otimes V_2$ is not a quotient of $V_1\times V_2$. For example, if both $V_1$ and $V_2$ have dimension $3$, the former has dimension $9$ while the latter only has dimension $6$. – Mariano Suárez-Alvarez Nov 3 '11 at 14:52
Agreed with Mariano. What perhaps IAmBrianDawkins meant is that $V_1\otimes V_2$ is a quotient of the free $\mathbb{Z}$-module on $\left\{m\otimes n:m,n\in M,N\right\}$. The ideal we mod has the nice quality that it precisely makes the dual basis...a basis (i.e. it adds in bilinearity). – Alex Youcis Nov 3 '11 at 17:04
Thank you, Mariano and Alex. Much appreciated. – Adam Saltz Nov 3 '11 at 18:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.