Self-Calibration Under the Cayley Framework

Abstract

The Cayley framework here is meant to tackle the vision problems under the infinite Cayley transformation (ICT), its main advantage lies in its numerical stability. In this work, the stratified self-calibration under the Cayley framework is investigated. It is well known that the main difficulty of the stratified self-calibration in multiple view geometry is to upgrade a projective reconstruction to an affine one, in other words, to estimate the unknown 3-vector of the plane at infinity, called the normal vector. To our knowledge, without any prior knowledge about the scene or the camera motion, the only available constraint on a moving camera with constant intrinsic parameters is the well-known Modulus Constraint in the literature. Do other kinds of constraints exist? If yes, what they are? How could they be used? In this work, such questions will be systematically investigated under the Cayley framework. Our key contributions include: 1. The original projective expression of the ICT is simplified and a new projective expression is derived to make the upgrade easier from a projective reconstruction to a metric reconstruction. 2. The constraints on the normal vector are systematically investigated. For two views, two constraints on the normal vector are derived; one of them is the well-known modulus constraint, while the other is a new inequality constraint. There are only these two constraints for two views. For three views, besides the constraints for two views, two groups of new constraints are derived and each of them contains three constraints. In other words, there are 12 constraints in total for three views. 3. Based on our projective expression and these constraints, a stratified Cayley algorithm and a total Cayley algorithm are proposed for the metric reconstruction from images. It is experimentally shown that they both improve significantly the numerical stability of the classical algorithms. Compared with the global optimal algorithm under the infinite homography framework, the Cayley algorithms have comparable calibration accuracy, but substantially reduce the computational load.

Appendices

Appendix A: Abbreviations and Notations

Abbreviations
ICT	Infinite Cayley transformation
ZEC	Zero-Eigenvalue constraint
CEC	Conjugate-Eigenvalue constraint
3-VRC	3-View row constraint
3-VCC	3-View column constraint
S-CA	Stratified Cayley algorithm
S-CAref	Stratified Cayley algorithm with the refinement step
T-CA	Total Cayley algorithm
POLLref	Pollefeys’ algorithm with the refinement step
ADQref	Absolute dual quadric based calibration with the refinement step
CHANgoa	Global optimal algorithm of Chandrakar et al.

Notations
$\text{ R}$	Rotation matrix
$[\mathbf{w}]_\times$	Skew-symmetric matrix defined by a 3-vector $\mathbf{w}$
$\text{ K}$	Intrinsic parameter matrix of camera
$\omega$	Image of the absolute conic
$\omega ^*$	Dual of $\omega$
$\text{ H}$	Homography of a plane between two views
$\text{ H}_\infty$	Infinite homography between two views
$\text{ C}_\infty$	ICT between two views
$\text{ Q}_\infty ^*$	Absolute dual quadric
$|\cdot |$	Determinant of a matrix
$\text{ tr}(\cdot )$	Trace of a matrix
$\text{ adj}(\cdot )$	Adjoint of a matrix

Appendix B: Cayley Representation of Rotation Matrix

Before introducing the Cayley representation, we first briefly review the Rodrigues representation of a rotation matrix. Let $\mathbf{t} \in R^{3}$ and ${\hat{\mathbf{t}}}=\mathbf{t}/\left\Vert \mathbf{t} \right\Vert,$ then the rotation matrix $\text{ R}$ with axis ${\hat{\mathbf{t}}}$ and angle $\Vert \mathbf{t}\Vert$ can be expressed as (Hartley and Zisserman 2000)

$$\begin{aligned} \text{ R}=e^{[\mathbf{t}]_\times }=\text{ I}+\frac{\sin (\left\Vert \mathbf{t} \right\Vert)}{\left\Vert \mathbf{t} \right\Vert}[\mathbf{t}]_\times +\frac{1-\cos (\left\Vert \mathbf{t} \right\Vert)}{\left\Vert \mathbf{t} \right\Vert^{2}}[\mathbf{t}]_\times ^{2} . \end{aligned}$$

(53)

The 3-vector $\mathbf{t}$ in (53) is called the Rodrigues representation of rotation matrix $\text{ R}.$ Note that in general case, the Rodrigues representation is a multiple-to-one representation, for example $\mathbf{t}=(2n\pi +\alpha )\mathbf{v}$ represents the same rotation matrix, where $\mathbf{v}$ is a unit vector and $n=1, 2,\ldots .$ If let

$$\begin{aligned} \mathbf{t}\in S\equiv \{\mathbf{t} |\;t_3 >0, \left\Vert \mathbf{t} \right\Vert\le \pi \}\cup \{\mathbf{t} |\;t_3 <0, \left\Vert \mathbf{t} \right\Vert<\pi \}, \end{aligned}$$

(54)

then the Rodrigues representation is of one-to-one (Hartley and Kahl 2009), i.e. (54) defines a one-to-one mapping between $S$ and the 3D rotation matrix group G(R). We always assume $\mathbf{t}\in S$ to ensure the uniqueness of Rodrigues representation.

Let $\text{ G}_\pi (\text{ R})=\{\text{ R}|\,\text{ angle(R)}= \pi \}$ be the set of rotation matrices with the angle $\pi .$ Then, the Cayley transformation (3) is a one-to-one mapping between $\text{ G}(\text{ R})\backslash \text{ G}_\pi (\text{ R})$ and the 3D linear space $\text{ L}(\text{ W})$ defined by $3\times 3$ symmetric matrices under matrix addition and scalar multiplication. And thus, the 3-vector $\mathbf{w}$ is a representation of rotation matrix in $\text{ G}(\text{ R})\backslash \text{ G}_\pi (\text{ R}),$ called the Cayley representation.

For $\text{ R}\in \text{ G}_\pi (\text{ R}),$ its eigenvalues must be $\{1,-1,-1\}.$ It is easy to verify $\text{ R}=\text{2}\mathbf{vv}^{\mathrm{T}}-\text{ I}$ by the Rodrigues representation, where $\mathbf{v}$ is the axis of $\text{ R}.$ And the following equation holds:

$$\begin{aligned} \text{ R}=\mathop {\lim }\limits _{s\rightarrow +\infty } \frac{\text{ I}-[s\mathbf{v}]_\times }{\text{ I}+[s\mathbf{v}]_\times }. \end{aligned}$$

(55)

This is because for an arbitrary $\mathbf{w}\in \mathcal R ^{3},$ by a direct computation there is

$$\begin{aligned}&(\text{ I}+[\mathbf{w}]_\times )^{-1}=\frac{1}{1+\left\Vert \mathbf{w} \right\Vert^{2}}\\&\quad \times \left( {{\begin{array}{l@{\quad }l@{\quad }l} {1+w_1^2 }&{w_1 w_2 +w_3 }&{w_1 w_3 -w_2 } \\ {w_1 w_2 -w_3 }&{1+w_2^2 }&{w_2 w_3 +w_1 } \\ {w_1 w_3 +w_2 }&{w_2 w_3 -w_1 }&{1+w_3^2 } \\ \end{array} }} \right). \end{aligned}$$

Setting $\mathbf{w}=s\mathbf{v}$ in the above equation,

$$\begin{aligned}&(\text{ I}+[s\mathbf{v}]_\times )^{-1}=\frac{1}{1+s^{2}}\\&\quad \times \left( {{\begin{array}{l@{\quad }l@{\quad }l} {1+s^{2}v_1^2 }&{s^{2}v_1 v_2 +sv_3 }&{s^{2}v_1 v_3 -sv_2 } \\ {s^{2}v_1 v_2 -sv_3 }&{1+s^{2}v_2^2 }&{s^{2}v_2 v_3 +sv_1 } \\ {s^{2}v_1 v_3 +sv_2 }&{s^{2}v_2 v_3 -sv_1 }&{1+s^{2}v_3^2 } \\ \end{array} }} \right). \end{aligned}$$

Then, there must be

$$\begin{aligned} \mathop {\lim }\limits _{s\rightarrow +\infty } (\text{ I}+[s\mathbf{v}]_\times )^{-1}=\left( {{\begin{array}{l@{\quad }l@{\quad }l} {v_1^2 }&{v_1 v_2 }&{v_1 v_3 } \\ {v_1 v_2 }&{v_2^2 }&{v_2 v_3 } \\ {v_1 v_3 }&{v_2 v_3 }&{v_3^2 } \\ \end{array} }} \right)=\mathbf{vv}^{\mathrm{T}}. \end{aligned}$$

Hence

$$\begin{aligned} \mathop {\lim }\limits _{s\rightarrow +\infty } \frac{\text{ I}-[s\mathbf{v}]_\times }{\text{ I}+[s\mathbf{v}]_\times }&= \mathop {\lim }\limits _{s\rightarrow +\infty } (2(\text{ I}+[s\mathbf{v}]_\times )^{-1}-\text{ I})\\&= 2\mathop {\lim }\limits _{s\rightarrow +\infty } (\text{ I}+[s\mathbf{v}]_\times )^{-1}-\text{ I}=2\mathbf{vv}^{\mathrm{T}}\!-\!\text{ I}. \end{aligned}$$

Therefore, the rotation matrix $\text{ R}$ with angle $\pi$ and axis $\mathbf{v}$ is the limit of the Cayley transformation $\varphi ([s\mathbf{v}]_\times )$ at $s\rightarrow +\infty ,$ corresponding to the infinity on the direction $\mathbf{v}.$

By the above discussions, the Cayley transformation gives a one-to-one mapping between the 3D rotation matrix group $\text{ G}(\text{ R})$ and the 3D projective space.

The following proposition shows the relationship of the Cayley representations of two rotation matrices and their product.

Proposition 12

Assume $\text{ R}_1 ,\;\text{ R}_2 , \text{ R}_1 \text{ R}_2 \in \text{ G}(\text{ R})\backslash \text{ G}_\pi (\text{ R})$ and $\mathbf{w}_1 , \mathbf{w}_2$ and $\mathbf{w}_3$ are respectively the Cayley representation of $\text{ R}_1 ,\;\text{ R}_2$ and $\text{ R}_1 \text{ R}_2 .$ Then

$$\begin{aligned} \mathbf{w}_3 =\frac{1}{1-\mathbf{w}_1^{\mathrm{T}} \mathbf{w}_2 }(\mathbf{w}_1 +\mathbf{w}_2 -(\mathbf{w}_1\times \mathbf{w}_2 )). \end{aligned}$$

(56)

This equation can be verified by direct computation.

For the relationship between the Cayley and Rodrigues representations of rotation matrix, we have the following proposition.

Proposition 13

Assume w and t are respectively the Cayley representation and the Rodrigues representation of rotation matrix $\text{ R} \, in \, \text{ G}(\text{ R})\backslash \text{ G}_\pi (\text{ R}), then$

$$\begin{aligned} \mathbf{w}=-\frac{\tan (\left\Vert {\mathbf{t}/2} \right\Vert)}{\left\Vert {\mathbf{t}/2} \right\Vert}(\mathbf{t}/2). \end{aligned}$$

(57)

Proof

From $e^{[\mathbf{t}]_\times }=\text{ R}=\dfrac{\text{ I}-[\mathbf{w }]_\times }{\text{ I}+[\mathbf{w }]_\times },$ we have

$$\begin{aligned}{}[\mathbf{w}]_\times =\frac{\text{ I}-e^{[\mathbf{t}]_\times }}{\text{ I}+e^{[\mathbf{t}]_\times }}. \end{aligned}$$

(58)

By $e^{[\mathbf{t}]_\times }=e^{[\mathbf{t}/2]_\times }e^{[\mathbf{t}/2]_\times }$ and $(e^{[\mathbf{t}/2]_\times })^{-1}=e^{-[\mathbf{t}/2]_\times },$

$$\begin{aligned} \text{ I}-e^{[\mathbf{t}]_\times }&= -\left( {e^{[\mathbf{t}/2]_\times }-e^{-[\mathbf{t}/2]_\times }} \right)e^{[\mathbf{t}/2]_\times }\\ \frac{1}{\text{ I}+e^{[\mathbf{t}]_\times }}&= \frac{e^{-[\mathbf{t}/2]_\times }}{e^{[\mathbf{t}/2]_\times }+e^{-[\mathbf{t}/2]_\times }}. \end{aligned}$$

Then, by substituting them into (58)

$$\begin{aligned}{}[\mathbf{w}]_\times =-\frac{e^{[\mathbf{t}/2]_\times }-e^{-[\mathbf{t}/2]_\times }}{e^{[\mathbf{t}/2]_\times }+e^{-[\mathbf{t}/2]_\times }}. \end{aligned}$$

(59)

From (53), we can obtain

$$\begin{aligned} e^{[\mathbf{t}/2]_\times }-e^{-[\mathbf{t}/2]_\times }=2\frac{\sin (\left\Vert {\mathbf{t}/2} \right\Vert)}{\left\Vert {\mathbf{t}/2} \right\Vert}[\mathbf{t}/2]_\times , \end{aligned}$$

(60)

$$\begin{aligned} e^{[\mathbf{t}/2]_\times }+e^{-[\mathbf{t}/2]_\times }=2\left( {\text{ I}+\frac{1-\cos (\left\Vert {\mathbf{t}/2} \right\Vert)}{\left\Vert {\mathbf{t}/2} \right\Vert^{2}}[\mathbf{t}/2]_\times ^2 } \right). \end{aligned}$$

(61)

By (61) and the power series $(1+x)^{-1}\!=\!\sum _{n=0}^\infty {(-1)^{n}x^{n}} (|x|<1),$ we have

$$\begin{aligned} \frac{1}{e^{[\mathbf{t}/2]_\times }+e^{-[\mathbf{t}/2]_\times }}&= \frac{1}{2}\left( {\text{ I}+\frac{1-\cos (\left\Vert {\mathbf{t}/2} \right\Vert)}{\left\Vert {\mathbf{t}/2} \right\Vert^{2}}[\mathbf{t}/2]_\times ^2 } \right)^{-1}\\ \nonumber&= \frac{1}{2}\sum _{n=0}^\infty {(-1)^{n}} \frac{(1-\cos (\left\Vert {\mathbf{t}/2} \right\Vert))^{n}}{\left\Vert {\mathbf{t}/2} \right\Vert^{2n}}[\mathbf{t}/2]_\times ^{2n}. \end{aligned}$$

(62)

Then, by substituting (60) and (62) into (59), we have

$$\begin{aligned}{}[\mathbf{w}]_\times \!=\!-\frac{\sin (\left\Vert {\mathbf{t}/2} \right\Vert)}{\left\Vert {\mathbf{t}/2} \right\Vert}\!\sum _{n=0}^\infty {(1\!-\!\cos (\left\Vert {\mathbf{t}/2} \right\Vert))^{n}} \frac{(-1)^{n}[\mathbf{t}/2]_\times ^{2n+1} }{\left\Vert {\mathbf{t}/2} \right\Vert^{2n}}. \nonumber \\ \end{aligned}$$

(63)

By the identity equation $[\mathbf{t}/2]_\times ^{2n+1} =(-1)^{n}\left\Vert {\mathbf{t}/2} \right\Vert^{2n}[\mathbf{t}/2]_\times ,$ (63) becomes

$$\begin{aligned}{}[\mathbf{w}]_\times&= -\frac{\sin (\left\Vert {\mathbf{t}/2} \right\Vert)}{\left\Vert {\mathbf{t}/2} \right\Vert}\sum _{n=0}^\infty {(1-\cos (\left\Vert {\mathbf{t}/2} \right\Vert))^{n}} [\mathbf{t}/2]_\times \\&= -\frac{1}{\left\Vert {\mathbf{t}/2} \right\Vert}\cdot \frac{\sin (\left\Vert {\mathbf{t}/2} \right\Vert)}{1-(1-\cos (\left\Vert {\mathbf{t}/2} \right\Vert))}[\mathbf{t}/2]_\times \\&= -\frac{\tan (\left\Vert {\mathbf{t}/2} \right\Vert)}{\left\Vert {\mathbf{t}/2} \right\Vert}[\mathbf{t}/2]_{\times } . \end{aligned}$$

Hence, Proposition 13 holds. $\square$

By Proposition 13, we have ${\hat{\mathbf{{t}}}}=-{\hat{\mathbf{{w}}}}$ and $\left\Vert \mathbf{t} \right\Vert=2\arctan (\left\Vert \mathbf{w} \right\Vert).$ Thus, the angle and axis of the rotation matrix corresponding to $\mathbf{w}$ is respectively $2\arctan (\left\Vert \mathbf{w} \right\Vert)$ and $-{\hat{\mathbf{{w}}}}.$ Besides, by $\left\Vert \mathbf{t} \right\Vert=2\arctan (\left\Vert \mathbf{w} \right\Vert)$ there should be $\left\Vert \mathbf{w} \right\Vert\rightarrow +\infty$ when $\left\Vert \mathbf{t} \right\Vert\rightarrow \pi ,$ this indicates once again that the rotation matrix $\text{ R}$ with the angle $\left\Vert \mathbf{t} \right\Vert=\pi$ and the axis ${\hat{\mathbf{{t}}}}$ is the limit of the Cayley transformation $\phi ([s{\hat{\mathbf{{t}}}}]_\times )$ at $s\rightarrow +\infty ,$ corresponding to the infinity on the direction ${\hat{\mathbf{{t}}}}.$

Appendix C: Proof of Proposition 2

For $\mathbf{w}_2 \in \text{ span}(\mathbf{w}_1 ),$ there must exist a scalar $s$ such that $\mathbf{w}_2 =s\mathbf{w}_1 ,$ then

$$\begin{aligned} \text{ C}_{\infty 2} =\text{ K}[\mathbf{w}_2 ]_\times \text{ K}^{-1}=s_1 \text{ K}[\mathbf{w}_1 ]_\times \text{ K}^{-1}=s_1 \text{ C}_{\infty 1}. \end{aligned}$$

Thus

$$\begin{aligned} \text{ C}_{\infty 1} \omega ^{*}+\omega ^{*}\text{ C}_{\infty 1}^{\mathrm{T}} =0\;\Leftrightarrow \;\text{ C}_{\infty 2} \omega ^{*}+\omega ^{*}\text{ C}_{\infty 2}^{\mathrm{T}} =0. \end{aligned}$$

So, Proposition 2(a) holds. The proof of Proposition 2(b) can be done by the following three steps:

• (1) If $\mathbf{w}\in \text{ span}(\mathbf{w}_1 ,\;\mathbf{w}_2 ),$ i.e. there exist two scalars $s_1 , s_2$ such that $\mathbf{w}\in s_1 \mathbf{w}_1 +s_2 \mathbf{w}_2 ,$ then

  $$\begin{aligned} \text{ C}_\infty =\text{ K}[s_1 \mathbf{w}_1 +s_2 \mathbf{w}_2 ]_\times \text{ K}^{-1}=s_1 \text{ C}_{\infty 1} +s_2 \text{ C}_{\infty 2} . \end{aligned}$$

  Thus

  $$\begin{aligned} \text{ C}_\infty \omega ^*+\omega ^*\text{ C}_\infty ^{\mathrm{T}}&= s_1(\text{ C}_{\infty 1} \omega ^*+\omega ^*\text{ C}_{\infty 1}^{\mathrm{T}} )\\&+s_2 (\text{ C}_{\infty 2} \omega ^*+\omega ^*\text{ C}_{\infty 2}^{\mathrm{T}} ). \end{aligned}$$

  Therefore, Proposition 2(b) holds in this case.

• (2) If $\mathbf{w}\in \text{ span}(\mathbf{w}_1 \times \mathbf{w}_2 )$ where $\mathbf{w}_1 \times \mathbf{w}_2$ is the cross product of $\mathbf{w}_1$ and $\mathbf{w}_2 ,$ then there exists a scalar s such that $\mathbf{w}=s\mathbf{w}_1 \times \mathbf{w}_2 .$ From

  $$\begin{aligned} \text{[C}_{\infty 1} ,\text{ C}_{\infty 2} ]&\triangleq \text{ C}_{\infty 1} \text{ C}_{\infty 2} -\text{ C}_{\infty 2} \text{ C}_{\infty 1}\\&= \text{ K([}\mathbf{w}_1 ]_\times \text{[}\mathbf{w}_2 ]_\times -\text{[}\mathbf{w}_2 ]_\times \text{[}\mathbf{w}_1 ]_\times )\text{ K}^{-1}\\&= \text{ K[}\mathbf{w}_1 \times \mathbf{w}_2 ]_\times \text{ K}^{-1}, \end{aligned}$$

  we have $\text{ C}_\infty =s[\text{ C}_{\infty 1}, \text{ C}_{\infty \text{2}}]$ and

  $$\begin{aligned}&\text{ C}_\infty \omega ^*+\omega ^*\text{ C}_\infty ^{\mathrm{T}} =s([\text{ C}_{\infty 1} ,\text{ C}_{\infty 2} ]\omega ^*\!+\!\omega ^ *[\text{ C}_{\infty 1} \text{,C}_{\infty 2} ]^{T})\\&\quad =s\left( \text{ C}_{\infty 2} (\text{ C}_{\infty 1} \omega ^ *+\omega ^*\text{ C}_{\infty 1}^{\mathrm{T}} )+(\text{ C}_{\infty 1} \omega ^*\right.\\&\qquad +\omega ^*\text{ C}_{\infty 1}^{\mathrm{T}}) \text{ C}_{\infty 2}^{\mathrm{T}} \\&\qquad \left. -\text{ C}_{\infty 1} (\text{ C}_{\infty 2} \omega ^*+\omega ^*\text{ C}_{\infty 2}^{\mathrm{T}} )-(\text{ C}_{\infty 2} \omega ^*\right.\\&\qquad \left.+\omega ^*\text{ C}_{\infty 2}^{\mathrm{T}} )\text{ C}_{\infty 1}^{\mathrm{T}} \right). \end{aligned}$$

  Therefore, Proposition 2(b) holds in this case.

• (3) For an arbitrary $\mathbf{w}\in \mathcal R ^{3},$ there must be $s_1 , \text{ s}_2 , \text{ s}_3$ such that

  $$\begin{aligned} \mathbf{w}=s_1 \mathbf{w}_1 +s_2 \mathbf{w}_2 +s_3 (\mathbf{w}_1 \times \mathbf{w}_2 ). \end{aligned}$$

  Thus

  $$\begin{aligned} \text{ C}_\infty =s_1 \text{ C}_{\infty 1} +s_2 \text{ C}_{\infty 2} +s_3 \text{[C}_{\infty 1} ,\text{ C}_{\infty 2} ]. \end{aligned}$$

  By the steps (1) and (2), Proposition 2(b) holds for the general case.

Appendix D: Proof of the Inequality (30)

For the normal vector, there must have

$$\begin{aligned} \text{ H}_{\infty ij} =\left| {\sigma _i \text{ H}_{ij} -\mathbf{e}_{ij} \mathbf{x}_i^{\mathrm{T}} } \right|^{-1/3}(\sigma _i \text{ H}_{ij} -\mathbf{e}_{ij} \mathbf{x}_i^{\mathrm{T}} ). \end{aligned}$$

So, by (26) in Sect. 4

$$\begin{aligned} \sigma _i \text{ H}_{ij} -\mathbf{e}_{ij} \mathbf{x}_i^{\mathrm{T}} =\left( {\frac{\sigma _i \mu _{ji} \left| {\text{ H}_j } \right|}{\mu _{ij} \left| {\text{ H}_i } \right|}} \right)\text{ H}_{\infty ij}. \end{aligned}$$

Similarly

$$\begin{aligned} \sigma _j \text{ H}_{ji} -\mathbf{e}_{ji} \mathbf{x}_j^{\mathrm{T}} =\left( {\frac{\sigma _j \mu _{ij} \left| {\text{ H}_i } \right|}{\mu _{ji} \left| {\text{ H}_j } \right|}} \right)\text{ H}_{\infty ji}. \end{aligned}$$

Thus, the ICT’s expression (22) in Sect. 4 can be rewritten as

$$\begin{aligned} \text{ C}_{\infty ij}&= \frac{\delta _{ij} }{\mu _{ij} \mu _{ji} \left| {\text{ H}_i } \right|\left| {\text{ H}_j } \right|}((\sigma _j \left| {\text{ H}_i } \right|^{ 2}\mu _{ij}^3 )\text{ H}_{\infty ji} \nonumber \\&-(\sigma _i \left| {\text{ H}_j } \right|^{ 2}\mu _{ji}^3 )\text{ H}_{\infty ij} ). \end{aligned}$$

(64)

On the other hand

$$\begin{aligned}&\text{ C}_{\infty ij} =\text{ K}\phi (\text{ R}_{ij} )\text{ K}^{-1},\;\text{ H}_{\infty ij} =\text{ KR}_{ij} \text{ K}^{-1},\;\;\nonumber \\&\text{ H}_{\infty ji} =\text{ KR}_{ij}^{-1} \text{ K}^{-1}, \end{aligned}$$

(65)

here $\text{ R}_{ij}$ is the rotation matrix of the view pair $\left\{ {i,j} \right\}$ and $\phi (\text{ R}_{ij} )=\frac{\text{ I}-\text{ R}_{ij} }{\text{ I}+\text{ R}_{ij} }$ is the Cayley transformation of $\text{ R}_{ij} .$ By substituting (65) into (64), we can obtain

$$\begin{aligned} \phi (\text{ R}_{ij} )&= \frac{\delta _{ij} }{\mu _{ij} \mu _{ji} \left| {\text{ H}_i } \right|\left| {\text{ H}_j } \right|}((\sigma _j \left| {\text{ H}_i } \right|^{ 2}\mu _{ij}^3 )\text{ R}_{ij}^{-1} \nonumber \\&-(\sigma _i \left| {\text{ H}_j } \right|^{ 2}\mu _{ji}^3 )\text{ R}_{ij} ). \end{aligned}$$

(66)

Let $\{1,\;e^{i\theta },e^{-i\theta }\}$ be eigenvalues of $\text{ R}_{ij}$ and $\{\mathbf{u}, \mathbf{v}, {\bar{\mathbf{v}}}\}$ the corresponding eigenvectors. Then, eigenvalues of $\phi (\text{ R}_{ij} )$ are $\{\phi (1), \phi (e^{i\theta }), \phi (e^{-i\theta })\}=\{0, -i\tan (\theta /2), i\tan (\theta /2)\}$ and the corresponding eigenvectors still are $\{\mathbf{u}, \mathbf{v},{\bar{\mathbf{v}}}\}.$ And thus, from (66) we have

$$\begin{aligned} \frac{\delta _{ij} }{\mu _{ij} \mu _{ji} \left| {\text{ H}_i } \right|\left| {\text{ H}_j } \right|}(\sigma _j \left| {\text{ H}_i } \right|^{ 2}\mu _{ij}^3 -\sigma _i \left| {\text{ H}_j } \right|^{ 2}\mu _{ji}^3 )=0 \end{aligned}$$

(67)

$$\begin{aligned}&\frac{\delta _{ij} }{\mu _{ij} \mu _{ji} \left| {\text{ H}_i } \right|\left| {\text{ H}_j } \right|}(\sigma _j \left| {\text{ H}_i } \right|^{ 2}\mu _{ij}^3 e^{-i\theta }-\sigma _i \left| {\text{ H}_j } \right|^{ 2}\mu _{ji}^3 e^{i\theta })\nonumber \\&\quad =-i\tan (\theta /2). \end{aligned}$$

(68)

Since $\frac{\delta _{ij} }{\mu _{ij} \mu _{ji} \left| {\text{ H}_i } \right|\left| {\text{ H}_j } \right|}\ne 0,$ by (67) we obtain again the ZEC

$$\begin{aligned} \sigma _j \left| {\text{ H}_i } \right|^{ 2}\mu _{ij}^3 =\sigma _i \left| {\text{ H}_j } \right|^{ 2}\mu _{ji}^{3}. \end{aligned}$$

(69)

By substituting (69) into (68) and since $\delta _{ij} = \frac{\mu _{ij} }{\mu _{ji} (\sigma _i \mu _{ji} |\text{ H}_{ij}|+\mu _{ij}^2 )},$

$$\begin{aligned} \frac{\sigma _i \mu _{ji} \left| {\text{ H}_j } \right|}{\sigma _i \mu _{ji} \left| {\text{ H}_j } \right|+\mu _{ij}^2 \left| {\text{ H}_i } \right|}(e^{-i\theta }-e^{i\theta })=-i\tan (\theta /2). \end{aligned}$$

Then

$$\begin{aligned} \frac{\sigma _i \mu _{ji} \left| {\text{ H}_j } \right|}{\sigma _i \mu _{ji} \left| {\text{ H}_j } \right|+\mu _{ij}^2 \left| {\text{ H}_i } \right|}=-\frac{i\tan (\theta /2)}{e^{-i\theta }-e^{i\theta }}=\frac{1}{4\cos ^{2}(\theta /2)}>0 . \end{aligned}$$

Or equivalently

$$\begin{aligned} (\sigma _i \mu _{ji} \left| {\text{ H}_j } \right| )(\sigma _i \mu _{ji} \left| {\text{ H}_j } \right|+\mu _{ij}^2 \left| {\text{ H}_i } \right| )>0 . \end{aligned}$$

Therefore, the inequality (30) holds.