Let V be a real n-dimensional vector space.

Show that V $\otimes _\mathbb{R}$ $\mathbb{C}$ is isomorphic to V + iV.

Note that V $\otimes _\mathbb{R}$ $\mathbb{C}$ is a real vector space and is called the complexification of V. It is also the tensor product of two real vector spaces, V and $\mathbb{C}$.

V + iV is a vector space over the complex numbers.

Is this the isomorphic map? $\phi$( v $\otimes$ (a+ib) )= av + ibv

If not, what would be an isomorphic map?

  • 3
    $\begingroup$ I think you really mean $\;V\otimes_{\Bbb R}\Bbb C\;$ = the tensor product over the reals, and not the direct sum, which is no complexification. $\endgroup$ – DonAntonio Oct 26 '13 at 21:08

The $\phi$ you have given will work. The only question is: How to prove it?

First of all, make sure you understand why $\phi$ is well-defined. I can't give any advice here, since I don't know exactly what definitions you are working from, but the key point is to make sure that, if $\sum_k v_k \otimes (a_k + ib_k) = \sum_k v_k' \otimes (a_k' + ib_k')$, then $\phi$ maps both expressions to the same thing.

(It is a very common mistake for people who are new to tensor products to check things like this only on simple tensors, which is not adequate to show that $\phi$ truly exists.)

Next, you want to make sure that $\phi$ is really a map of vector spaces over $\mathbb{C}$. That is, it should preserve addition: $\phi(x+y)$ must equal $\phi(x)+\phi(y)$ for every $x,y \in V\otimes_{\mathbb{R}} \mathbb{C}$—this includes all sums, not just simple tensors. And furthermore, for any $c\in \mathbb{C}$, and any $x \in V\otimes_{\mathbb{R}} \mathbb{C}$, it must be true that $\phi(cx) = c\phi(x)$. Third warning: check all sums, not just simple tensors!

Finally, there are several approaches you could take to finish the proof from this point:

  • You could show that $\phi$ is both injective and surjective.
  • You could show that $\phi$ is injective, and that $\operatorname{dim}_{\mathbb{C}} V\otimes_{\mathbb{R}} \mathbb{C} = \operatorname{dim}_{\mathbb{C}} (V+iV)$.
  • You could show that $\phi$ is surjective, and that $\operatorname{dim}_{\mathbb{C}} V\otimes_{\mathbb{R}} \mathbb{C} = \operatorname{dim}_{\mathbb{C}} (V+iV)$.
  • You could explicitly construct a linear map $\psi: (V+iV) \to V\otimes_{\mathbb{R}} \mathbb{C}$ such that $\psi\circ\phi$ is the identity on $V\otimes_{\mathbb{R}} \mathbb{C}$, and $\phi\circ\psi$ is the identity on $(V+iV)$.

Any of these is good enough to prove that $\phi$ is an isomorphism—though the second and third approaches only work for finite-dimensional spaces, and you should make sure that you don't confuse real dimension with complex dimension.

(My opinion is that the fourth option will be the most instructive, but also the slowest.)

It will probably also be important throughout to make sure that you have a very clear definition of $V+iV$. The use of "$+$" here could be a little confusing, but as long as you can clearly explain your notion (both to yourself and to whoever will be looking at your proof), it should be fine.

(As has been pointed out, you can avoid much of the work here by dimension-counting. But I think that your approach is much better, for two reasons: first of all, it generalizes to infinite-dimensional spaces, and secondly, it is very important not just here, but in mathematics in general, to know what certain isomorphisms are, not just to know that they exist.)


Hint: Prove that if $\;\{v_1,...,v_n\}\;$ is a basis for $\;V_{\Bbb R}\;$ , and since $\;\{1,i\}\;$ is a basis for $\;\Bbb C_{\Bbb R}\;$ ,then

$$\{v_1,...,v_n, v_1i,...,v_ni\}\;\;\text{ is a basis for}\;\;V\otimes_{\Bbb R}\Bbb C$$

  • $\begingroup$ Yes i already know that. If dimV = n and dimW = m, then the dimension of the tensor product is nm. In this case the dimension of the complexification of V is 2n. How would I use that? thanks $\endgroup$ – sarah Oct 26 '13 at 21:16
  • $\begingroup$ I don't get it: don't you see what a basis for $\;iV\;$ is? And thus what a basis for $\;V\oplus iV\;$ is? $\endgroup$ – DonAntonio Oct 26 '13 at 22:04
  • $\begingroup$ Or even much... much ...easier: any two vector spaces with the same dimension over the same field are isomorphic...! $\endgroup$ – DonAntonio Oct 26 '13 at 22:05
  • $\begingroup$ isn't the tensor product a vector space over reals while V + iV is a vector space over complex? $\endgroup$ – sarah Oct 28 '13 at 18:11
  • $\begingroup$ $\;V\oplus iV\;$ is a real space, which can be made a complex space, too. $\endgroup$ – DonAntonio Oct 28 '13 at 19:32

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