Question regarding Tensor product

Question

Let $\ T: V\times W \rightarrow \mathbb V\otimes W$ be a map defined as $\ T(v,w) = v\otimes\ w$ where $\ v \in V,w\in W $. Then T is bilinear. Further if $\ (v_1,\ldots,v_n)\ and\;\ (w_1,\ldots,w_m)$ are bases of V and W respectively, then $(v_1\otimes w_1,\dots,v_n \otimes w_m)$ form a basis of $ V \otimes W$. I get the bilinearity part. But what does that kind of a basis say about the dimension of $V\otimes W$. Is the dimension $max(m,n)$, or am I missing something here?

Answer 1

First, you're mixing up the universal property of the tensor product. Let $V,W$ be real vector spaces. Here's the correct statement: If $T:V\times W\to X$ is a bilinear map of real vector spaces, there exists a unique linear map $\tilde{T}:V\otimes W\to X$ which makes the following diagram commute: $$\begin{matrix} V\times W & \xrightarrow{T} & X \\ \downarrow & \nearrow \tilde{T} & \\ V\otimes W & & \end{matrix}.$$

Second, you're correct that bases $\{v_i\}$ of $V$ and $\{w_j\}$ of $W$ induce a basis $\{v_i\otimes w_j\}$ of $V\otimes W$. So to see the dimension of $V\otimes W$, just count! How many basis elements $\{v_i\otimes w_j\}_{i=1,\ldots,\dim V;j=1,\ldots,\dim W}$ are there? Answer: $(\dim V)(\dim W)$.

Answer 2

If $V$ is $n$ dimensional, and $W$ is $m$ dimensional, then $V \otimes W$ has dimension $nm$.

See http://mathworld.wolfram.com/VectorSpaceTensorProduct.html