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

Trace of a Tensor

The trace of a tensor is a scalar that obeys the following rules: For any tensor \(\mbf{T}\) and \(\mbf{S}\) and any vectors \(\mbf{a}\) and \(\mbf{b}\),

\[\begin{align} \text{tr}(\mbf{T} + \mbf{S}) = \text{tr}\mbf{T} +\text{tr}\mbf{S}, \\ \text{tr}(\alpha\mbf{T}) = \alpha \text{tr}\mbf{T}, \\ \text{tr}(\mbf{a}\mbf{b}) = \mbf{a}\cdot\mbf{b}. \end{align}\]

In terms of tensor components, using Eq. (2.12.7),

\[\begin{align} \text{tr}\mbf{T} = \text{tr}(\mathit{T_{ij}}\mbf{e_i}\mbf{e_j}) = \mathit{T_{ij}}\text{tr}(\mbf{e_i}\mbf{e_j}) = \mathit{T_{ij}}\mbf{e_i}\mbf{e_j} = \mathit{T_{ij}}\mathit{\delta_{ij}} = \mathit{T_{ii}}. \end{align}\]

That is,

\[\begin{align} \text{tr}\mbf{T} = \mathit{T_{11}} + \mathit{T_{22}} + \mathit{T_{33}} = \text{sum of diagonal elements}. \end{align}\]

It is, therefore, obvious that

\[\begin{align} \text{tr}\mbf{T}^T = \text{tr}\mbf{T} \end{align}\]

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