Why $Vdiag(-1,-1,1)V^T$ represents a rotation through $180^\circ$ about $u_3$ provided that some matrix $E=Udiag(1,1,0)V^T$

Question

Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision at page 259: 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?

Answer 1

$\mathtt{VW}^T\mathtt W^T\mathtt V^T$ 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.