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.

• Isn't the result that you're trying to prove a part of the last property on your list? – Omnomnomnom May 22 at 1:55
• It is - I am being asked to prove the last property, as noted in the final part of my question. – dnfost May 22 at 2:16
• Ok I didn't understand what you meant by that last sentence but now it makes sense I guess – Omnomnomnom May 22 at 2:17
• 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 $A$)? – Omnomnomnom May 22 at 2:20
• I believe what you described is correct. It seems like the dot product with order 2 and 1 tensors would just be matrix multiplication. – 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 $$u \cdot (v \otimes w) = u^T(vw^T) = (u^Tv)w^T = (u \cdot v)\, w^T.$$