One of the main obstacles in understanding the tensor product is that, unlike many other algebraic structures, you cannot really get hold of its element structure. This confuses many beginners. The universal property tells us that this doesn't really matter and it helps us to construct linear maps on the tensor product, but it takes a while to appreciate this level of abstraction. So what about constructing the tensor product in a different way? Here is a suggestion. I would like to hear what you think about it.

Let $V,W$ be vector spaces over a field $K$. Let us assume that $V$ is finite-dimensional. Define the tensor product $V \otimes W := \hom(V^*,W)$. This is a vector space by construction. Its elements are just linear maps $V^* \to W$. If $v \in V$, $w \in W$, we define the pure tensor $v \otimes w \in V \otimes W$ by $(v \otimes w)(\omega)=\omega(v) \cdot w$. Observe that $(v+v') \otimes w=v \otimes w + v' \otimes w$ and $(\lambda v) \otimes w = \lambda(v \otimes w)$, likewise for the other variable, i.e. $V \times W \to V \otimes W$ is a bilinear map.

Any element $\alpha \in V^* \otimes W$ is a sum of pure tensors: If $e_1,\dotsc,e_n$ is a basis of $V$, then $\alpha = \sum_i e_i \otimes \alpha(e_i^*)$. (Because the right hand side maps $e_i^*$ to $\alpha(e_i^*)$.)

Universal property: If $b : V \times W \to U$ is a bilinear map, there is a unique linear map $f : V \otimes W \to U$ such that $f(v \otimes w)=b(v,w)$ for all $(v,w) \in V \times W$. Proof. Uniqueness follows since every element is a sum of pure tensors. For existence, we just let $f(\alpha)=\sum_i b(e_i,\alpha(e_i))$. Then one easily checks $f(e_j \otimes w)=b(e_j,w)$ and therefore $f(v \otimes w)=b(v,w)$. $\checkmark$

The universal property shows in particular that the tensor product of finite-dimensional vector spaces is symmetric: There is a unique isomorphism $V \otimes W \cong W \otimes V$ mapping $v \otimes w \mapsto w \otimes v$. [Of course, lots of other nice properties of the tensor product also follow from the universal property.]

If $V$ is not assumed to be finite-dimensional, we define $V \otimes W$ as the subspace of $\hom(V^*,W)$ which is generated by the pure tensors $v \otimes w$ as defined above. (Thus, if $V$ is finite-dimensional, there will be no difference.) The universal property follows immediately.

  • $\begingroup$ Martin, this is pretty nice. From an algebraic point of view, I think it definitely is more enjoyable. Quotienting out by the equivalence relations that we would like (so that tensor products are in some ways generalizations of multiplication) definitely feels like "cheating" to some degree. I do have one question: aside from aesthetics, why would this be more natural? Would you be willing to write an MSE blog post on this matter? I'd love to see a more expansive look at defining tensor products in this manner. $\endgroup$ Sep 28, 2014 at 21:55
  • $\begingroup$ this construction is well known $\endgroup$
    – janmarqz
    Jun 7, 2018 at 22:49

1 Answer 1


I would like to hear what you think about it.

I don't like it. You have to choose a basis to prove the universal property.

But recently I was forced to study a certain representation and learn how it can be decomposed as a 'tensor product,' which led me to learn about the 'tensor product of Hilbert spaces, $\mathcal{H}_1 \otimes \mathcal{H}_2$.'

And now suddenly it seems that there is no convenient way I can do this without thinking of a certain subspace of $B(\mathcal{H}_1^*, \mathcal{H}_2)$ equipped with a certain inner product. (The above notation is for bounded linear operators.)


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