In particular, how is it that dimension of Cartesian product is a sum of dimensions of underlying vector spaces, while Tensor product, often defined as a quotient of Cartesian product, has dimension which is a product of dimensions of underlying vector spaces.

Can this be easily visualized and explained in the case of RxR Cartesian product?

How do the requirements of bilinearity and linearity cause a 'switch' from (x1,x2) to x1*x2?

  • $\begingroup$ unlike in the cartesian product where (u,v) + (w,x) = (u+w,v+x), the addition operation on elements of the tensor product is not defined component-wise $\endgroup$
    – user785085
    Oct 17, 2021 at 13:09

2 Answers 2


The tensor product of two vector spaces is not a quotient of the Cartesian product of those spaces. It is a quotient of the free vector space with basis the cartesian product. That is, $V \otimes W$ is a quotient of an enormous infinite dimensional vector space. A vector space with basis $$\{x_\alpha \ | \ \alpha \in V \times W\},$$ so there is one basis element for each element of $V \times W$ (hence enormous).

If you fix bases $\{v_i\}$ and $\{w_j\}$ of $V$ and $W$ then, because of the relations that one quotients by, the tensor product has as basis those $\overline{x_\alpha}$ (where the overline means the element of the quotient represented by $x_\alpha$) with $\alpha = (v_i, w_j)$ for some $i, j$. There are $\dim V$ choices for $v_i$ and $\dim W$ choices for $w_j$ hence there are $(\dim V)(\dim W)$ choices for $\alpha$. Thus $\dim(V \otimes W) = (\dim V)(\dim W)$.

On the other hand the cartesian product $V \times W$ has basis $$\{(v_i, 0), (0, w_j) \ | \ \text{for all} \ i, j\}$$ and there are $\dim V + \dim W$ elements in that set so $\dim(V \times W) = \dim V + \dim W$.


Given two vector spaces over the same ring $K$: $(V_{1},+_{1},*_{1})$, $(V_{2},+_{2},*_{2}).$ The Cartesian product of the two vector spaces is the new vector space $(V_{1}\times V_{2},+,*)$ with the new operations $(x,y)+(z,w)=(x+_{1}z,y+_{2}w),$ $\alpha*(x,y)=(\alpha*_{1}x,\alpha*_{2}y)$ for every $x,z\in V_{1},\;\;$ $y,w\in V_{2}$ and $\alpha\in K.$

The tensor product of the vector spaces $V_{1}$ and $V_{2}$ is the set of all linear (in both arguments) applications (mappings) from $V_{1}\times V_{2}$ onto $K;$ which means $V_{1}\otimes V_{2}=\{T:V_{1}\times V_{2}\mapsto K,\;\;T(\alpha x+\beta z,y)=\alpha*T(x,y)+\beta*T(z,y),\;\;T(x, \rho y+\sigma w)=\rho*T(x,y)+\sigma*T(x,w) \}.$

The difference between Cartesian and Tensor product of two vector spaces is that the elements of the cartesian product are vectors and in the tensor product are linear applications (mappings), this last are vectors as well but these ones applied onto elements of $V_{1}\times V_{2}$ gives a $K-$number.

  • 2
    $\begingroup$ Thank you for the great explanation! $\endgroup$
    – Obriareos
    Mar 24, 2020 at 22:32
  • 1
    $\begingroup$ I assume "linear application" means "linear map" right? $\endgroup$ Jul 29, 2020 at 5:09
  • $\begingroup$ @BrianHaak indeed $\endgroup$ Jul 29, 2020 at 10:47

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