I am working through some lecture notes which cover an introduction to tensors and the tensor product between vector spaces.

The setup is that $K$ is a field, $S$ is a set and $KS = \oplus_{s \in S}K$ denotes the free vector space built upon the basis $S$. We let $R$ denote the subspace of $KS$ spanned by all "relations" of the following three types:

$$ (v_1 + v_2) \otimes w - (v_1 \otimes w + v_2 \otimes w) $$ $$ v \otimes (w_1 + w_2) - (v \otimes w_1 + v\otimes w_2) $$ $$ v \otimes (\lambda \cdot w) - (\lambda \cdot v) \otimes w - \lambda \cdot (v \otimes w). $$

The tensor product is defined then by the quotient space $V \otimes W: = KS/R$. What I don't understand is how you can use "relations" to span a subspace of the free vector space $KS$. What does the subspace $R$ really look like? In my head I imagine three different types of subspaces which correspond to all of the relations given above but I know this can't be the correct reasoning, since $R$ will only be one subspace.

Maybe my question is more general to application to constructing a tensor product; how can subspaces be defined using equivalence relations?

  • $\begingroup$ Is your quote correct? The construction I know of $V\otimes W$ does not come from any $S$, but from $S=V\times W$ specifically. $\endgroup$
    – user562983
    Sep 10, 2021 at 7:13
  • $\begingroup$ It's just an abuse of terminology. What it means is that $R$ is spanned by free space vectors given by one of these three forms. If you use vectors from the original space to represent elements of the quotient then you have to make them obey relations obtained by equating these three expressions to $0$. In other words, linear combinations of such vectors are exactly the elements of the kernel of the quotient map. $\endgroup$
    – Conifold
    Sep 10, 2021 at 8:29

1 Answer 1


First of all, the relations are not correct. They should instead be the following: $$(v_1+v_2)\otimes w - (v_1 \otimes w + v_1\otimes w),$$ $$v \otimes (w_1 + w_2) - (v \otimes w_1 + v\otimes w_2),$$ $$v \otimes (\lambda \cdot w) - \lambda \cdot (v \otimes w),$$ $$(\lambda \cdot v) \otimes w - \lambda \cdot (v \otimes w),$$ for all $v,v_1,v_2\in V$, all $w,w_1,w_2\in W$ and all $\lambda\in K$, and where $v\otimes w:=(v,w)\in KS$ with $S:=V\times W$.

As for your actual question; the elements of the vector space $KS$, where $S=V\times W$, are precisely the finite sums of the form $$\sum_{(v,w)\in V\times W}\lambda_{(v,w)}(v,w).$$ That is, they are simply formal finite $K$-linear combinations of pairs $(v,w)\in V\times W$. With the convention that $v\otimes w:=(v,w)\in KS$ we see that the relations above are elements of this vector space $KS$. These elements span some linear subspace $R$, and so we can take the quotient $KS/R$.

For a more tangible example that is similar but a bit contrived, consider the vector space $W=K[X]$ and the subspace $R$ spanned by the functions $$\lambda X+\mu,\qquad \lambda X^2+\mu,$$ for all $\lambda\in K$. Together these span a subspace $R$ of $W$ that you can describe very explicitly. You can also describe the quotient $W/R$ very explicitly.

The vector space $KS$ with $S=V\times W$ is of course much larger and more abstract. I would recommend not to try to imagine what $R$ looks like in this case. In fact I would recommend to not give this whole construction much thought. As I see it, the construction is an existence proof for $V\otimes W$, where $V\otimes W$ is a vector space defined by a universal property. This property is by far the most relevant property of the tensor product, and is for most purposes the only thing you need to know about a tensor product.


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