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

Representation of tensors as a sum of dyadics products

Say \(\bs{T}\) is a tensor and \(T_{ij}\) are its components w.r.t. to the basis \(\hat{\bs{E}}_i\), i.e.,

\[\begin{align} T_{ij}=\hat{\bs{E}}_i\cdot(\bs{T} \hat{\bs{E}}_j) \end{align}\]

then it can be shown that

\[\begin{align} \bs{T}&=T_{ij}\hat{\bs{E}}_{i}\otimes \hat{\bs{E}}_{j} \tag{DyadicBasis} \end{align}\]

Let

\[\begin{align} \bs{T}\bs{c}&=\bs{d}\\ (T_{ij}\hat{\bs{E}}_{i}\otimes \hat{\bs{E}}_{j})\bs{c}&=\bs{e} \end{align}\]

We are going to find the components of \(\bs{e}\) and \(\bs{d}\) and show that they are one and the same. Since \(\bs{c}\) is arbitrary this goes to prove that \(\bs{T}\) and \(T_{ij}\hat{\bs{E}}_{i}\otimes \hat{\bs{E}}_{j}\) are one and the same.

Components of \(\bs{e}\) \(\begin{align} \bs{e}&=(T_{ij}\hat{\bs{E}}_{i}\otimes \hat{\bs{E}}_{j})\bs{c}\\ &= (T_{ij}\hat{\bs{E}}_{i}\otimes \hat{\bs{E}}_{j})(c_k\hat{\bs{E}}_k)\\ &= T_{ij} c_k(\hat{\bs{E}}_{i}\otimes \hat{\bs{E}}_{j})\hat{\bs{E}}_k\\ &= T_{ij} c_k(\hat{\bs{E}}_{i}( \hat{\bs{E}}_{j}\cdot\hat{\bs{E}}_k))\\ &= T_{ij} c_k(\hat{\bs{E}}_{i}\delta_{jk})\\ &= T_{ij} c_j\hat{\bs{E}}_{i} \end{align}\)

From the last equation it follows that

\[\begin{align} \bs{e}\cdot\hat{\bs{E}}_j&=T_{ik}c_k\hat{\bs{E}}_i\cdot\hat{\bs{E}}_j\\ e_j&=T_{ik}c_k\delta_{ij}\\ e_j&=T_{jk}c_k\label{eq:eComponents}\\ \end{align}\]

Components of \(\bs{d}\)

From the definition of the matrix definition of \(\bs{T}\), which is \([\bs{T}]\) it follows that

\[\begin{align} \left[\bs{T}\right]\left[\bs{c}\right]&=\left[\bs{d}\right]\label{eq:MatrixFrom} \end{align}\]

where \(\begin{align} \left[\bs{c}\right]_{i}&=c_{i}=\bs{c}\cdot \hat{\bs{E}}_i\\ \left[\bs{d}\right]_{j}&=d_{j}=\bs{d}\cdot \hat{\bs{E}}_j \end{align}\)

In Einstein notation $\eqref{eq:MatrixFrom}$ reads,

\[\begin{align} d_i&=T_{ij}c_j\label{eq:dComponents} \end{align}\]

If follows from $\eqref{eq:eComponents}$ and $\eqref{eq:dComponents}$ that \(\bs{e}\) and \(\bs{d}\) are the same, from which it follows that that \(\bs{T}\) and \(T_{ij}\hat{\bs{E}}_{i}\otimes \hat{\bs{E}}_{j}\) are one and the same.

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