Consider first the trivial case when one of the spaces, \(V_i\) say, in a tensor product space, \(V_1\otimes\dots\otimes V_r\), is zero. Then any \(r\)-linear function out of \(V_1\times\dots\times V_r\) must be zero and so, as the image of the \(r\)-linear function \(\iota:V_1\times\dots\times V_r\mapto V_1\otimes\dots\otimes V_r\) generates the whole tensor product space, \(\dim(V_1\otimes\dots\otimes V_r)=0\) in this case. More generally, in the case that none of the spaces are zero, observe first that the dimension of \(V_1\otimes\dots\otimes V_r\) is the dimension of the dual space, \((V_1\otimes\dots\otimes V_r)^*\), and as already discussed, \(\mathcal{L}(V_1,\dots,V_r;K)\cong(V_1\otimes\dots\otimes V_r)^*\). Now suppose that a basis for the \(i\)th space, \(V_i\), is \(\{e_1^{(i)},\dots,e_{n_i}^{(i)}\}\), that is, \(\dim V_i=n_i\), for \(i=1,\dots,r\). Then for any \(r\)-linear form, \(f\), we have
\begin{equation}
f\left(\sum_{i_1}c_{(1)}^{i_1}e_{i_1}^{(1)},\dots,\sum_{i_1}c_{(r)}^{i_r}e_{i_r}^{(r)}\right)=\sum_{i_1,\dots,i_r}c_{(1)}^{i_1}\cdots c_{(r)}^{i_r}f(e_{i_1}^{(1)},\dots,e_{i_r}^{(r)}),
\end{equation}
where the \(c_{(j)}^{i_j}\) are arbitrary scalars. That is, \(f\) is uniquely specified by the \(n_1\cdots n_r\) scalars \(f(e_{i_1}^{(1)},\dots,e_{i_r}^{(r)})\). So defining the \(n_1\cdots n_r\), clearly linearly independent, \(r\)-linear forms, \(\phi_{i_1\dots i_r}\), such that, \(\phi_{i_i\dots i_r}(e_{j_1}^{(1)},\dots,e_{j_r}^{(r)})=\delta_{i_1j_1}\dots\delta_{i_rj_r}\), that is, \(\phi_{i_1\dots i_r}=\alpha_{i_1}^{(1)}(\cdot)\cdots\alpha_{i_r}^{(r)}(\cdot)\), where \(\{\alpha_1^{(i)},\dots,\alpha_{n_i}^{(i)}\}\) is the dual basis of \(V_i^*\), we see that any \(r\)-linear form can be expressed as linear combination of the \(\phi_{i_i\dots i_r}\) and so they form a basis for \(\mathcal{L}(V_1,\dots,V_r;K)\) which therefore has dimension \(n_1\cdots n_r\). That is,
\begin{equation}
\dim(V_1\otimes\dots\otimes V_r)=\dim V_1\cdots\dim V_r.
\end{equation}
It is also then clear that the \(n_1\cdots n_r\) pure tensors \(\{e_{i_1}^{(1)}\otimes\dots\otimes e_{i_r}^{(r)}\}\) form a basis for \(V_1\otimes\dots\otimes V_r\).

A nice application of the tensor product machinery is to the construction of the complexification of a vector space \(V\) over \(\RR\). As a vector space over \(\RR\), \(\CC\) is a two dimensional vector space with basis \(\{1,i\}\). Suppose \(\{e_1,\dots,e_n\}\) is a basis for the \(n\)-dimensional space \(V\). Then we can form the tensor product space \(\CC\otimes V\) over \(\RR\). As a real vector space this has basis \(\{1\otimes e_1,\dots,1\otimes e_n,i\otimes e_1,\dots,i\otimes e_n\}\) but we can define scalar multiplication by complex numbers simply as \(z(z’\otimes v)=(zz’)\otimes v\). We should now demonstrate that \(\CC\otimes V\cong V_\CC\) where \(V_\CC\) is as defined in Realification and Complexification. Consider the map \(\phi:V_\CC\mapto\CC\otimes V\) defined by \(\phi(v,v’)=1\otimes v+i\otimes v’\). This is clearly linear over \(\RR\). That it’s also linear over \(\CC\) follows since \(\phi(i(v,v’))=\phi(-v’,v)=-1\otimes v’+i\otimes v=i(1\otimes v+i\otimes v’)=i\phi(v,v’)\). To verify that this is an isomorphism , we’ll construct the inverse map. The bilinear map \(\CC\times V\mapto V_\CC\) defined by \((z,v)\mapsto z(v,0)\) induces by virtue of the universal mapping property a linear map we’ll call \(\phi’:\CC\otimes V\mapto V_\CC\) given by \(\phi'(z\otimes v)=z(v,0)\) on pure tensors. But this map is obviously also \(\CC\)-linear and since, \(\phi\circ\phi'(z\otimes v)=\phi(z(v,0))=z\phi(v,0)=z(1\otimes v)=z\otimes v\), we see that it is the inverse of \(\phi\).

Let us also note at this point that we have the obvious isomorphisms,
\begin{equation}
K\otimes V\cong V\cong V\otimes K,
\end{equation}
when \(V\) is a vector space over \(K\).