Originally Posted by

**Sogan** hmmm. I really don't see it. Let us assume that the vs above are isomorphic.

The dimensions are: dim($\displaystyle (V\times W) \otimes U)=dim(V \otimes U)*dim(W \otimes U)$ how do you seperate the LHS?

What are you talking about? In general as has been pointed out $\displaystyle \displaystyle \dim\left(V\otimes W\right)=\dim\left(V\right)\dim\left(W\right)$ and thus in particular

$\displaystyle \begin{aligned}\dim\left(\left(V\oplus W\right)\otimes U\right) &=\dim\left(V\oplus W\right)\dim\left(U\right)\\ &=\left(\dim\left(V\right)+\dim\left(W\right)\righ t)\dim\left(U\right)\\ &=\dim(V)\dim(U)+\dim(W)\dim(U)\end{aligned}$

And this is certainly not equal to $\displaystyle \dim(V)\dim(W)\dim(U)^2$ as you seemed to indicate. Now, back to your original question if you don't want to appeal to a dimensional argument and explicitly want to write down an isomorphism try taking the obvious one. Namely, if $\displaystyle V$ has basis $\displaystyle \{v_1,\cdots,v_n\}$ and $\displaystyle W$ has basis $\displaystyle \{w_1,\cdots,w_m\}$ then try considering the unique linear homomorphism $\displaystyle T:V\otimes W\to\text{Hom}\left(V,W\right)$ such that $\displaystyle v_i\otimes w_j\mapsto T_{i,j}$ where $\displaystyle T_{i,j}:V\to W$ is the unique element of $\displaystyle \text{Hom}\left(V,W\right)$ with $\displaystyle v_i\mapsto w_j$ and $\displaystyle v_k\mapsto \bold{0},\text{ }k\in[n]-\{i\}$. This is evidently an isomorphism since it's linear and a bijection between bases.

P.S. There are multiple ways of thinking about the tensor product. I was taught initially, before I was given the definition you mentioned, to think of $\displaystyle V\otimes W$ as $\displaystyle \text{Bil}\left(V,W\right)^\ast$ where $\displaystyle \text{Bil}(V,W)$ is the space of bilinear forms on $\displaystyle V\oplus W$.