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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?

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Your definition $T(v,w)=v\otimes w$ does not match $T:V\times W\to \mathbb R$ except if $V$ and $W$ are both $\mathbb R^1$. –  Henning Makholm Apr 14 '12 at 1:50
    
The 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.. –  anon Apr 14 '12 at 1:55
    
@Henning My apologies for the error. I have fixed it. –  Vishesh Apr 14 '12 at 1:59
    
I 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)=<v,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? –  Vishesh Apr 14 '12 at 2:21
    
Looks 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...) –  anon Apr 14 '12 at 2:24
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up vote 2 down vote accepted

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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Thanks 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. –  Vishesh Apr 14 '12 at 2:04
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If $V$ is $n$ dimensional, and $W$ is $m$ dimensional, then $V \otimes W$ has dimension $nm$.

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

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