I'm looking at Sergei Winitzki's Linear Algebra via Exterior Products, and he has a question on tensor products.

Firstly we construct the real vector space $\Bbb R^3 \otimes_\Bbb R \Bbb C$ which is 'basically' vectors of the form $v_1 \otimes 1+v_2 \otimes i$ with $v_1,v_2 \in \Bbb R^3$. As a real vector space it is $6$ dimensional.

Then he defines multiplication by a complex number on $\Bbb R^3 \otimes_\Bbb R \Bbb C$ for $\lambda \in \Bbb C$ by

$\lambda(v \otimes z)=v \otimes( \lambda z)$. Then asks what is the dimension of this as a $\Bbb C$ vector space.

I have a feeling that this is now isomorphic to $\Bbb R^3$. The two previously linearly independent vectors $v \otimes 1$ and $v \otimes i$ are now multiples of each other.

I feel like in some sense you're now treating $\Bbb C$ as a vector space over itself in the tensor product now so the dimension is $3\times1$.

I'm not sure if this is the right way to think of these things though.

  • 1
    $\begingroup$ It looks like $\mathbb{R}^3$ in that it has the "same basis". It is a vector space over $\mathbb{C}$ with basis any basis of $\mathbb{R}^3$ as a real vector space. Hence, $\dim_{\mathbb{C}} =3$. $\endgroup$ – Orest Bucicovschi Oct 20 '14 at 18:00
  • $\begingroup$ Really it doesn't look like $\mathbb{R}^3$ but rather $\mathbb{C}^3$. $\endgroup$ – Oscar Cunningham Oct 21 '14 at 7:18
  • $\begingroup$ $\Bbb C^3$ over $\Bbb C$. Yeah that would actually make more sense :) Thanks! $\endgroup$ – snulty Oct 21 '14 at 8:49
  • $\begingroup$ In that case they should be isomorphic by dimension. I may try write down an isomorphism $\endgroup$ – snulty Oct 21 '14 at 8:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.