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Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision at page 259: enter image description here Here, some matrix $E$(called essential matrix) can be decomposed as $$E=Udiag(1,1,0)V^T=\begin{bmatrix}u_1&u_2&u_3\end{bmatrix}diag(1,1,0) \begin{bmatrix}v_1^T\\v_2^T\\v_3^T\end{bmatrix}$$ To my understanding, rotating through $180^\circ$ about $u_3$(the unit direction vector joining the two camera centers) can be writen as $$2u_3u_3^T-I$$ accoring to Rodrigues' rotation formula, but $Vdiag(-1,-1,1)V^T=2v_3v_3^T-I$. Some details must be left out, who can help?

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  • $\begingroup$ $\operatorname{diag}(-1,-1,1)$ is a rotation about the third ($z$) axis through an angle of $\pi$. Conjugating with $V$ changes to a coordinate system in which the columns of $V$ are the coordinate axes, i.e., it’s a rotation about $v_3$. $\endgroup$
    – amd
    Dec 25, 2017 at 2:25
  • $\begingroup$ I can figure out it is a rotation about $v_3$, but the book said it is a rotation about $u_3$ which is the line joining two camera centers, referring to Multiple View Geometry in Computer Vision at page 259 $\endgroup$
    – Finley
    Dec 25, 2017 at 2:38
  • $\begingroup$ No, the book says that it’s a rotation about the line joining the camera centers. The second camera center isn’t at $\mathbf t$; it’s at $-\mathtt R^T\mathbf t$. $\endgroup$
    – amd
    Dec 27, 2017 at 10:33

1 Answer 1

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$\mathtt{VW}^T\mathtt W^T\mathtt V^T$ isn’t a rotation about $\operatorname{span}\{\mathbf u_3\}$. It’s a rotation about the line joining the camera centers. Although it’s true that $\mathbf t$ is a multiple of $\mathbf u_3$, that’s not where the second camera is located. It’s at $-\mathtt R^T\mathbf t$ (p.156).

The rotation axis of $\mathtt{VW}^T\mathtt W^T\mathtt V^T = \mathtt V\operatorname{diag}(-1,-1,1)\mathtt V^T$ is $\operatorname{span}\{\mathbf v_3\}$. From the SVD of $\mathtt E$ we can see that $\mathbf v_3$ is a right null vector of $\mathtt E$, so we have $$\mathtt E\mathbf v_3 = \mathtt R\,[\mathtt R^T\mathbf t]_\times\mathbf v_3 = 0 \implies [\mathtt R^T\mathbf t]_\times\mathbf v_3 = 0,$$ hence $\mathbf v_3$ is a multiple of $\mathtt R^T\mathbf t$ and $\operatorname{span}\{\mathbf v_3\}$ is the line joining the camera centers.

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  • $\begingroup$ Cool!. I originally thought nobody will answer my two posts, but you answered both, Thanks for taking your time on my trivial questions. $\endgroup$
    – Finley
    Dec 27, 2017 at 11:46
  • $\begingroup$ @Finley Keep asking. It’s good to see that you’re really trying to understand the material. $\endgroup$
    – amd
    Dec 27, 2017 at 18:01

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