I'm having a tough time trying to understand tensor product and how to compute it. I don't know what to do when I have $\alpha\in V^*$, $\vec v\in V$ and I want to compute $\alpha\otimes\vec v$ or $\vec v\otimes\alpha$... For example, if I have $\alpha = (e^3-4e^1)$ and $\vec v = (-2\vec e_1+\vec e_3)$, $\alpha\otimes\vec v$ and $\vec v\otimes\alpha$ look like what?

The definition that my professor has taught me is this: given $\alpha,\beta\in V^*$ and $\vec u,\vec v\in V$ we have $(\alpha\otimes\beta)(\vec u,\vec v)\doteq \alpha(\vec u)\beta(\vec v)$. Thank you very much, folks.

  • $\begingroup$ There are several definitions of the "tensor" product that will make $\alpha \times v$ look (at least superficially) very different. It would help us if you could write the definition that you're using, or at least point the textbook that you're using. $\endgroup$ Apr 23, 2017 at 22:33
  • $\begingroup$ I meant to write $\alpha \otimes v$, of course. Please remember that you can edit your question to add details. $\endgroup$ Apr 23, 2017 at 22:40
  • $\begingroup$ Thank you! I've forgotten to edit my question!! $\endgroup$ Apr 23, 2017 at 22:46
  • $\begingroup$ Why has somebody downvoted this question (I have upvoted it) ? Caue Evangelista is a newcomer, that has tried to her best to explain her problem. $\endgroup$
    – Jean Marie
    Apr 23, 2017 at 22:47
  • $\begingroup$ Thank very much for your kind words JeanMarie, I really don't understand this type of thing as well... $\endgroup$ Apr 23, 2017 at 22:51

2 Answers 2


The definition you wrote is for the tensor product of two linear functionals. You can consider the tensor product of a linear functional with a vector by "plugging in what makes sense", that is: $$(\alpha \otimes v)(w, \beta) \doteq \alpha(w)\beta(v),$$and so on. What you can do for now is write that as combination of "basic" tensors, using bilinearity of $\otimes$, as in: $$\begin{align} \alpha \otimes v &= (e^3 - 4e^1)\otimes(-2e_1+e_3)\\ &= -2 e^3\otimes e_1 + e^3\otimes e_3 + 8e^1\otimes e_1 - 4 e^1\otimes e_3.\end{align}$$If $f = f_i e^i$ and $w = w^j e_j$, you can see how $\alpha \otimes v$ acts explicitly by the above formula: $$(\alpha \otimes v)(w,f) = -2w^3f_1 + w^3f_3 + 8f^1w_1 - 4f^1w_3.$$

  • $\begingroup$ Thank you soo much Ivo! I thought earlier that maybe $\alpha$$\otimes$$\vec v$ is done this way but I was not sure! Seriously, thank you so much! My best wishes! $\endgroup$ Apr 23, 2017 at 22:56
  • $\begingroup$ By the way, you don't need to use all these dollars \$ when typing mathjax here. See my edit to your question. $\endgroup$
    – Ivo Terek
    Apr 23, 2017 at 22:57
  • $\begingroup$ Oh, it's much more easy to read in your way! Thanks! $\endgroup$ Apr 23, 2017 at 23:00

Probably you are using for aeV* and veV that aoxv is the bilinear form on VxV* where v acts on b* in V* by point evaluation - v (b*) = b*(v) (thus identifying V** with V so aoxv(w,b*) = a(w)b*(v) .This works as long as V is finite dimensional .Hope that helps .What book are you using ?

  • $\begingroup$ I'm using the Arfken's book "Mathematical Methods for Physicists" $\endgroup$ Apr 23, 2017 at 23:07

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