\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Dyadic product

The dyadic product between two tensor \(\bs{a}\) and \(\bs{b}\) as the tensor \(\bs{a}\otimes \bs{b}\), such that given any vector \(\bs{c}\)

\[\begin{align} (\bs{a}\otimes \bs{b})\bs{c}&=\bs{a}(\bs{b}\cdot\bs{c}) \end{align}\]

A tensor of the type \(\bs{a}\otimes \bs{b}\) is sometimes referred to as a dyad.

Properities of Dyads

Let \(\bs{A}\) a tensor. Then, \(\begin{align} \left(\bs{a}\otimes \bs{b}\right)^{\textsf{T}}&=\bs{b}\otimes \bs{a}\\ \left(\bs{a}\otimes \bs{b}\right) \bs{A}&= \bs{a}\otimes \left(\bs{A}^{\textsf{T}}\bs{b} \right)\\ \bs{A} \left(\bs{a}\otimes \bs{b}\right)&= \left(\bs{A}\bs{a} \right)\otimes \bs{b} \\ \end{align}\)

Proof.

Let \(\bs{c}\) be some arbitrary vector, then,

\[\begin{align} \left(\bs{a}\otimes \bs{b}\right) \bs{A}\bs{c}&=\bs{a}\left(\bs{b}\cdot \bs{A}\bs{c}\right)\\ &=\bs{a}\left(\bs{c}\cdot \bs{A}^{\textsf{T}}\bs{b}\right)\\ &=\bs{a}\left(\left(\bs{A}^{\textsf{T}}\bs{b}\right)\cdot \bs{c} \right)\\ &=\left(\bs{a}\otimes \left(\bs{A}^{\textsf{T}}\bs{b}\right)\right) \bs{c}\\ \end{align}\] \[\begin{align} \bs{A} \left(\bs{a}\otimes \bs{b}\right) \bs{c}&=\bs{A}\bs{a}\left(\bs{b}\cdot \bs{c}\right)\\ &=\left(\bs{A}\bs{a}\otimes \bs{b} \right) \bs{c}\\ \end{align}\]

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