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.

  • $\begingroup$ 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

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 $$ u \cdot (v \otimes w) = u^T(vw^T) = (u^Tv)w^T = (u \cdot v)\, w^T. $$

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