Epipolar Rectification by Singular Value Decomposition of Essential Matrix

Abstract

Image rectification is an important stage of applying a pair of projective transformations, or homographies, to a pair of stereo images, so that epipolar lines in the original images map to horizontally aligned lines in the rectified images. Considering that for some stereo rigs the intrinsic parameters of the cameras are known but their external parameters are unknown, in this paper, we present a novel method for stereo rectification based on the essential matrix which is derived from the fundamental matrix. Without any optimization process, closed-form analytical solutions of the projective transformations for epipolar rectification can be directly obtained by conducting SVD on the essential matrix. Experimental results show the proposed rectification method not only has higher efficiency and rectification precision, but also its scale invariance and robustness are superior to existing methods.

Appendix A

Proposition

For any non-zero vector t = [t_x, t_y, t_z]^T, the symmetric matrix $\mathrm{A}={\left[\mathbf{t}\right]}_{\times }{\left[\mathbf{t}\right]}_{\times}^{\mathrm{T}}$ has two equal eigenvalues which are equal to the square of the norm of vector t, and its third eigenvalue is 0.

Proof

Due to$\mathrm{A}\mathbf{t}=\Big({\left[\mathbf{t}\right]}_{\times}\left[\mathbf{t}\Big]{}_{\times}^{\mathrm{T}}\right)\mathbf{t}=0$, it can be concluded that zero is an eigenvalue of the symmetric matrix A as well as its eigenvector corresponding to zero is t. Furthermore, A can be written as

$$ \kern0.5em \mathbf{A}={\left[\mathbf{t}\right]}_{\times }{\left[\mathbf{t}\right]}_{\times}^{\mathrm{T}}=-{\left[\mathbf{t}\right]}_{\times }{\left[\mathbf{t}\right]}_{\times }=-\left[\begin{array}{ccc}\hfill -\left({\mathrm{t}}_{\mathrm{z}}^2+{\mathrm{t}}_{\mathrm{y}}^2\right)\hfill & \hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{y}}\hfill & \hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{z}}\hfill \\ {}\hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{y}}\hfill & \hfill -\left({\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{z}}^2\right)\hfill & \hfill {\mathrm{t}}_{\mathrm{y}}{\mathrm{t}}_{\mathrm{z}}\hfill \\ {}\hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{z}}\hfill & \hfill {\mathrm{t}}_{\mathrm{y}}{\mathrm{t}}_{\mathrm{z}}\hfill & \hfill -\left({\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{y}}^2\right)\hfill \end{array}\right] $$

Let λ ₁ = 0 , λ ₂ , λ ₃ are the three eigenvalues of A, it follows that

$$ {\displaystyle \begin{array}{l}\left|\uplambda \mathbf{I}-\mathbf{A}\right|=\left|\begin{array}{ccc}\hfill \uplambda -\left({\mathrm{t}}_{\mathrm{z}}^2+{\mathrm{t}}_{\mathrm{y}}^2\right)\hfill & \hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{y}}\hfill & \hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{z}}\hfill \\ {}\hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{y}}\hfill & \hfill \uplambda -\left({\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{z}}^2\right)\hfill & \hfill {\mathrm{t}}_{\mathrm{y}}{\mathrm{t}}_{\mathrm{z}}\hfill \\ {}\hfill {\mathrm{t}}_{\mathrm{x}}{\mathrm{t}}_{\mathrm{z}}\hfill & \hfill {\mathrm{t}}_{\mathrm{y}}{\mathrm{t}}_{\mathrm{z}}\hfill & \hfill \uplambda -\left({\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{y}}^2\right)\hfill \end{array}\right|\hfill \\ {}=\uplambda \left(\uplambda -{\uplambda}_2\right)\left(\uplambda -{\uplambda}_3\right)={\uplambda}^3-\left({\uplambda}_2+{\uplambda}_3\right){\uplambda}^2+{\uplambda}_2{\uplambda}_3\uplambda =0\hfill \end{array}} $$

Then we have that

$$ {\displaystyle \begin{array}{l}{\uplambda}_2+{\uplambda}_3={\mathrm{A}}_{11}+{\mathrm{A}}_{22}+{\mathrm{A}}_{33}=2\left({\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{y}}^2+{\mathrm{t}}_{\mathrm{z}}^2\right)\hfill \\ {}{\uplambda}_2{\uplambda}_3={\mathrm{t}}_{\mathrm{x}}^2{\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{y}}^2{\mathrm{t}}_{\mathrm{y}}^2+{\mathrm{t}}_{\mathrm{z}}^2{\mathrm{t}}_{\mathrm{z}}^2+2{\mathrm{t}}_{\mathrm{x}}^2{\mathrm{t}}_{\mathrm{y}}^2+2{\mathrm{t}}_{\mathrm{y}}^2{\mathrm{t}}_{\mathrm{z}}^2+2{\mathrm{t}}_{\mathrm{x}}^2{\mathrm{t}}_{\mathrm{z}}^2={\left({\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{y}}^2+{\mathrm{t}}_{\mathrm{z}}^2\right)}^2\hfill \end{array}} $$

It can be easily obtained from above two equations that

$$ {\uplambda}_2={\uplambda}_3={\mathrm{t}}_{\mathrm{x}}^2+{\mathrm{t}}_{\mathrm{y}}^2+{\mathrm{t}}_{\mathrm{z}}^2={\mathbf{t}}^2 $$