2
$\begingroup$

How would I prove that $u \cdot (v \otimes w) = (u \cdot v) \otimes w$ where $u, v,$ and $w$ are first-order tensors ?

The only concepts I've learned so far are properties of the dyadic product, which are:

$(u \otimes v) \neq (v \otimes u) \\(u \otimes v) \cdot w = u \otimes (v \cdot w) = u (v \cdot w) = (v \cdot w) u \\ u \otimes (\alpha v + \beta w) = \alpha u \otimes v + \beta u \otimes w \\ (u \otimes v) (w \otimes x) = (u \otimes x)(v \cdot w) \\ u \cdot (v \otimes w) = ( u \cdot v) \otimes w = (u \cdot v) w = w(u \cdot v)$

Using these, I've tried $ u \cdot ( v \otimes w ) = (v \otimes w) \cdot u $ from the commutative property of the dot product, $ \\ (v \otimes w) \cdot u = v \otimes (w \cdot u)$ from the second listed property of the dyadic product, $ v \otimes (w \cdot u) = v(w \cdot u) $ from the second listed property of the dyadic product, but I seem to go in circles after this point, eventually just arriving back where I started.

It doesn't look to me like there are more properties left to use other than the last, but that's what I'm trying to prove. I'd appreciate any advice or insight.

$\endgroup$
  • $\begingroup$ Isn't the result that you're trying to prove a part of the last property on your list? $\endgroup$ – Omnomnomnom May 22 at 1:55
  • $\begingroup$ It is - I am being asked to prove the last property, as noted in the final part of my question. $\endgroup$ – dnfost May 22 at 2:16
  • $\begingroup$ Ok I didn't understand what you meant by that last sentence but now it makes sense I guess $\endgroup$ – Omnomnomnom May 22 at 2:17
  • $\begingroup$ Are you given some kind of definition of a dot product? In particular, how are we supposed to compute the dot-product $u \cdot A$ where $u$ is order $1$ and $A$ is order $2$? Is it the same thing as the product $u^TA$ (dot product of $u$ with all columns of $A$)? $\endgroup$ – Omnomnomnom May 22 at 2:20
  • $\begingroup$ I believe what you described is correct. It seems like the dot product with order 2 and 1 tensors would just be matrix multiplication. $\endgroup$ – dnfost May 22 at 15:01
1
$\begingroup$

Assuming that my comment is correct: let $u,v,w$ denote the column vectors associated with the tensors in question (so that we can lean on matrix multiplication). We have $$ u \cdot (v \otimes w) = u^T(vw^T) = (u^Tv)w^T = (u \cdot v)\, w^T. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.