Suppose $V,W$ are vector spaces with bases $\{e_1,e_2,\dots ,e_m\}$ and $\{f_1,f_2,\dots ,f_n\}$. I know that $V\otimes W=F(V\times W)/H$, where $F(V\times W)$ is the free vector space on $V\times W$ and $H$ is the subspace generated by all elements satisfying certain relations such that bi-linearity holds in the quotient space. I'm writing $v\otimes w$ for the congruency class of the pair $(v,w)\in V\times W$. I can easily show that the pairs $e_i\otimes f_j$ span $V\otimes W$, but I'm having trouble showing linear independence i.e. that if $\sum_{i,j}\lambda_{ij}e_i\otimes f_j=0$ then for all $i,j$, $\lambda_{ij}=0$. Could anybody point me in the right direction? Thanks for any replies!

  • $\begingroup$ Try using the universal property of the tensor product: linear maps out of the tensor product $V \otimes W$ are bilinear maps out of $V \times W$. Note that the dimension of a finite dimensional vector space is the dimension of its dual. So use the universal property to figure out the dimension of $(V \otimes W)^{*}.$ $\endgroup$ – Siddharth Venkatesh May 13 '14 at 19:17
  • $\begingroup$ sorry but could you expand? I'm not able to piece this all together, and even if I could figure out the dimension of the dual space (I don't know what that is) then I'd have no idea where to go from there! $\endgroup$ – James Machin May 13 '14 at 20:14
  • $\begingroup$ I added an answer. $\endgroup$ – Siddharth Venkatesh May 13 '14 at 22:32

$\def\Hom{\mathrm{Hom}}$ So, since you already know that the product of the bases spans $V \otimes W$, you just need to show that the dimension of $V \otimes W$, which is the dimension of its dual, is the product of the dimension of $V$ and the dimension of $W$.

We first reduce the problem to showing that $$(V \otimes W)^{*} = \Hom_{k}(V \otimes W, k) \cong \Hom_{k}(V, \Hom_{k}(W, k)) = \Hom_{k}(V, W^{*}).$$

Note that if the above relationship holds, then the dimension of $V \otimes W$ is the dimension of the space of linear maps from $V$ to $W^{*}$, which is just the product of the dimensions of $V$ and $W^{*}$, and is hence the product of the dimensions of $V$ and $W$.

So, to show the above equalities, note that the leftmost and rightmost equalities are by definition of the dual. So we need to show that

$$\Hom_{k}(V \otimes W, k) \cong \Hom_{k}(V, \Hom_{k}(W, k)).$$

This is actually true if $k$ is any ring and $V$ and $W$ are any $k$-modules. This is the adjunction between $\Hom$ and tensor product. But let's go through the details for the vector space case.

Suppose $\phi$ is a linear map from $V \otimes W$ to $k$. We want to define $f(\phi)$ which takes a vector in $v$ and spits out a linear map from $W$ to $k$. To do so, we define

$$(f(\phi) (v)) (w) = \phi(v \otimes w).$$

You should check that this map is linear and well-defined. It is not too hard. We next construct an inverse map. Given $\psi$ which to each $v$ assigns a linear map from $W$ to $k$, we define $g(\psi)$ as a linear map from $V \otimes W$ to $k$ by first defining it on the simple tensors as $$g(\psi)(v \otimes w) = \psi(v)(w)$$ noting that this is bilinear in the $v$ and $w$ and then noting that bilinear maps extend to all of $V \otimes W$.

Again, $g$ can be checked to be linear and you can also easily see that $f$ and $g$ are inverses.

  • $\begingroup$ Ok thanks a lot! I've had to take a step back and start going through more basis, as my foundations haven't been good enough to understand some of this stuff. I'm hoping to come back and go through this later on though $\endgroup$ – James Machin May 14 '14 at 15:05

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