Prove ${\bf u}.{\bf Tv} = ({\bf u} \otimes {\bf v}) : {\bf T}$

Question

Example 8 (on Page 6) of these lecture notes appear to suggest that for any vectors $\bf u$ and $\bf v$ and tensor $\bf T$

$ {\bf u}.{\bf Tv} = ({\bf u} \otimes {\bf v}) : {\bf T} $

I'd appreciate help proving this.

Answer 1

To reframe this in terms of matrix multiplication, note that for column-vectors $x,y$ and matrices $A,B$ of the same shape, we have

• $x \cdot y = x^\top y$,
• $u \otimes v = uv^\top $,
• $A : B = \operatorname{Tr}(A^\top B)$.

With that, we can prove this result as a consequence of the cyclic property of trace. Or, if you prefer, we could use the slightly simpler result $\operatorname{Tr}(xy^\top) = x^\top y$ for column-vectors $x,y$. In any case, we have $$ (u \otimes v): T = \operatorname{Tr}((uv^\top)^\top T) = \operatorname{Tr}(v u^\top T) = \operatorname{Tr}(u^\top Tv) = u^\top (Tv) = u \cdot Tv $$ as desired.

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If we want to stick to dyadic notation, we might state the following: $T$ can necessarily be written as a sum $T = x_i \otimes y_i$ for some vectors $x_i,y_i$. It follows that $$ (u \otimes v) : T = (u \otimes v):(x_i \otimes y_i) = (u \cdot x_i)(v \cdot y_i),\\ u \cdot Tv = u \cdot (x_i \otimes y_i)v = (u \cdot x_i)(v \cdot y_i). $$ It follows that the two expressions are indeed equal.