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?
Question regarding Tensor product
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1Your definition $T(v,w)=v\otimes w$ does not match $T:V\times W\to \mathbb R$ except if $V$ and $W$ are both $\mathbb R^1$. – 2012-04-14
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0The basis is $\{v_i \otimes v_j\}_{i=1,\cdots, n;~ j=1,\cdots m}$. I don't see how you're indexing what you call the basis.. – 2012-04-14
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0@Henning My apologies for the error. I have fixed it. – 2012-04-14
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0I have something of an extended question.I hope its ok to ask that here rather than start another separate thread.Continuing on Neal's answer, my book offers a slight modification that there is a unique linear isomorphism **S** of $ V \otimes W $ onto $ L(V*,W) $ such that $ S(v \otimes w)(a)=
w $ where v and w are vectors as earlier and a is in $ V* $. My question is I have only ever seen dot product of 2 vectors and "a" here is a linear mapping Is the dot product here due the fact that V and V* are isomorphic(finite dimension only) or is there some other idea involved? – 2012-04-14 -
0Looks to me by $a\in V^\vee$ and $v\in V$ they mean $a(v)$ by $\langle v,a\rangle$ (in fact $\langle \varphi|v\rangle:=\varphi(v)$ is notation just introduced in something I'm reading...) – 2012-04-14
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0The notation can be confusing; an inner product is an example of a more general phenomenon of "pairing," where you take two elements of two vector spaces (or modules) and evaluate them against each other. If they're finite dimensional and you can prove this pairing is nondegenerate, it induces an isomorphism between the vector spaces. In fact, the inner product $\langle\cdot,\cdot\rangle$ may be regarded as a pairing between $V^*$ and $V$, which induces an isomorphism between them (when they're finite dimensional). – 2012-04-14
2 Answers
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: $(\dim V)(\dim W)$.
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0Thanks to everyone for those immediate answers. I just did not see the pairing involved in the basis which led me to the max(m,n) conclusion. – 2012-04-14
If $V$ is $n$ dimensional, and $W$ is $m$ dimensional, then $V \otimes W$ has dimension $nm$.
See http://mathworld.wolfram.com/VectorSpaceTensorProduct.html