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

Orthogonal Tensor

Definition A special orthogonal tensor is tensor \(\bs{Q}\) that satisfies the equation

\[\begin{align} \bs{Q}^{\textsf{T}}\bs{Q}&=\bs{I}\\ \textsf{det}(\bs{Q})&=1 \end{align}\]

The first equation implies that \(\bs{Q}\bs{Q}^{\textsf{T}}=\bs{I}\). The proof of this result is not simple, and hence omitted here.

Proposition An orthogonal tensor has a unit eigen value.[^1]

Proof

From the definition of the eigen value problem we have that

\[\begin{align} \bs{Q}\hat{\bs{v}}&=\lambda\hat{\bs{v}},\\ \end{align}\]

where \(\lambda\) is called the eigen values and \(\hat{\bs{v}}\) is the called the eigen vector. As denoted by the overhat, the vector \(\hat{\bs{v}}\) is of unit magnitude, i.e., \(\begin{align} \bs{v}\cdot \bs{v}=1 \end{align}\)

\[\begin{align} (\bs{Q}\hat{\bs{v}})\cdot (\bs{Q}\hat{\bs{v}}) &=(\lambda\hat{\bs{v}})\cdot (\lambda\hat{\bs{v}})\\ \hat{\bs{v}}\cdot (\bs{Q}^{\textsf{T}}\bs{Q}\hat{\bs{v}}) &=\lambda^2 \hat{\bs{v}}\cdot \hat{\bs{v}}\\ \hat{\bs{v}}\cdot (\bs{I}\hat{\bs{v}}) &=\lambda^2 \hat{\bs{v}}\cdot \hat{\bs{v}}\\ \hat{\bs{v}}\cdot \hat{\bs{v}} &=\lambda^2 \hat{\bs{v}}\cdot \hat{\bs{v}}\\ 1 &=\lambda^2 \hat{\bs{v}}\cdot \hat{\bs{v}}\\ 1 &=\lambda^2,\\ \end{align}\]

which implies that

\[\begin{equation} \lambda=\pm 1 \end{equation}\]

In the following we show that \(+1\) is definitely an eigen value of an orthogonal tensor. The eigen value \(\lambda\) is a root of the characteristic polynomial

\[\begin{equation} \textsf{det}(\bs{Q}-\lambda \bs{I})=0 \end{equation}\]

Since, \(\bs{Q}:\mathbb{E}^3 \mapsto \mathbb{E}^3\), the characteristic polynomial is of third order. So, it can have three real roots or one real root and two complex, conjugate roots.

Let us consider case one: The possibilities are \(\alpha=\pm 1\), \(\beta=\pm 1\), and \(\gamma =\pm 1\).We know that \(\textsf{det}(\bs{Q})=1\), thus the only possiblities all three positive, two negative one positive. (not allowed: two positive, one negative, all negative). Thus, there is atleast one eigen value that is \(+1\).

Let us consider case two, where there is a real root and two complex roots, \(\alpha,~a+i b,~a-i b\). We know that \(\textsf{det}(\bs{Q})=1\), thus

\[\begin{align} \alpha (a+ib) (a-i b)&=1\\ \alpha (a^2+b^2)&=1\\ \alpha \lVert a+ib \rVert^2&=1\\ \alpha &=1 \end{align}\]

[^1] If \(\lambda\) is an eigen value of a tensor, then it has to be a root of of a characteristic polynomial. However, are all roots of a characteristic polynomial eigen values?

[^2] The above derivation needs to be redone by taking into account the fact that the eigen values can be complex.

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