Is the tensor product of two vector space on $R$ isomorphic to $R^{d^2}$?

Question

Let $V$,$W$ be two fine-dimensional vector spaces over the field of real numbers $\mathbb{R}$. Assume the dimension of both spaces is d. Is there a unique isomorphism between $V \otimes W$ and $\mathbb{R}^{d^2}$?

ie, can we interpret elements of $V \otimes W$ as being vectors in a $\mathbb{R}^{d^2}$ space?

So on the easiest case scenario, is an element of $V \otimes W$ just a real number?

Answer 1

Assuming that $V$ is $n$-dimensional and that $W$ is $m$-dimensional, then $V \otimes W$ is $mn$-dimensional. Indeed, assuming that $V$ has a basis $\{e_i\}_{i=1}^n$ and that $W$ has a basis $\{f_j\}_{j=1}^m$, then you can show that $V\otimes W$ has a basis $\{e_i\otimes f_j\}_{i,j}$, and so it's $mn$-dimensional. See here. Your claim then follows from the fact that any $mn$-dimensional normed space (over $\mathbb{R}$) is isomorphic to $\mathbb{R}^{mn}$ (see van Neerven, J. (2022). Functional analysis (Vol. 201). Cambridge University Press, for instance).

Answer 2

Suppose $V$ and $W$ are finite-dimensional vector spaces over the field $\Bbb{R}$ of the real numbers saying

$\text{dim}(V) = n$ and $\text{dim}(W) = m \qquad n,m \in \Bbb{N}$

Their tensor-product has the dimension

$\text{dim}(V \otimes W) = n \cdot m$

as stated in S. Lang: Algebra, Revised third edition, p.609, Cor. 2.4.

Because two finite-dimensional vector spaces are isomorphic iff they have equal dimension there will be an isomorphism, but this isomorphism has not to be necessarily unique.

So, the vectors of the tensor-product can be regarded als $n\cdot m$ tuples of real numbers, but they are not real numbers for their own. The vector space structure is that of a tensor-product with its universal property.

Answer 3

Let $\{v_1,...,v_n\}$ and $\{w_1,...,w_m\}$ be bases for $V$ and $W$ respectively. The tensor product space $V\otimes W$ is a vector space with basis from objects (vectors in the product space) of form $v_i \otimes w_j$ for $1\leq i \leq n$, $1\leq j\leq m$. Thus dimension of $V\otimes W$ is $nm$.

To make things less abstract, if $V,W$ were space of column vectors with $n,m$ entries in $\mathbb{R}$, tensor product is the space spanned by $n\times m$ matrices $E_{ij}$ whose $ij$th enry is $1$ and else is $0$. Thus dimension is $nm$.

Now, I think the word you are looking for is canonical or natural, which to me it the one above, but the set of isomorphisms between two same dimensional vector spaces over the same fields is a group, and in case of $\mathbb{R}$ there is not just single such isomorphism. (e.g if $A$ is such isomorphism, so is $2A$).