Stuck in deriving decomposition of essential matrix(background in computer vision needed)

Question

Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 258 to 259. In particular Where $$W=\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1 \end{bmatrix},Z=\begin{bmatrix}0&1&0\\-1&0&0\\0&0&0 \end{bmatrix}$$ and $S,R$ are skew-symmetric matrix and rotation matrix of $3 \times 3$ respectively.$$S=[t]_\times=\begin{bmatrix}0&-t_3&t_2\\t_3&0&-t_1\\-t_2&t_1&0 \end{bmatrix}$$

Question1: Why $S=UZU^T$ if left null space of $S$ is the same as that of $E$?

Question2: How to show Frobenius norm of $S=UZU^T$ is $\sqrt2$ and as a result leading to $\| t\|_2=1$?

Answer 1

Proof to question1:

Having $E$ factorized as $E=Udiag(1,1,0)V^T$, we can derive $EE^T=Udiag(1,1,0)U^T$ . And observe that $EE^T=SR(SR)^T=SS^T=-S^2$ due to skew-symmetry of $S$, at this point we get $$Udiag(1,1,0)U^T=-S^2$$besides$$Z^2=\begin{bmatrix}0&1&0\\-1&0&0\\0&0&0 \end{bmatrix}^2 =-diag(1,1,0)$$ so $S^2=UZ^2U^T=UZU^TUZU^T$ which leads to $S=UZU^T$ according to definition of matrix power.

Answer 2

For your first question, we know from the given SVD that $\mathtt E$ restricted to the complement of its null space is length-preserving, therefore the restriction of $\mathtt S$ is, too. This means that it must be orthogonally similar to $\mathtt Z$ (result A4.1), i.e., $\mathtt S = \mathtt U'\mathtt Z\mathtt U'^T$ for some orthogonal matrix $\mathtt U'$. Based on the observation at the end of section 9.6.1, we can arrange for $\mathtt U'=\mathtt U$.

As to your second question, a straightforward calculation verifies the results: Using the facts that $\mathtt U$ is orthogonal and that the trace is invariant under conjugation, $$\|\mathtt S\|_F^2 = \operatorname{tr}(\mathtt S^T\mathtt S) = \operatorname{tr}(\mathtt{UZ}^T\mathtt U^T\mathtt{UZU}^T) = \operatorname{tr}(\mathtt{UZ}^T\mathtt{ZU}^T) = \operatorname{tr}(\mathtt Z^T\mathtt Z) = \operatorname{tr}\operatorname{diag}(1,1,0) = 2.$$ On the other hand, another direct calculation shows that the square of the Frobenius norm of $[\mathbf t]_\times$ is $2(t_x^2+t_y^2+t_z^2) = 2\|\mathbf t\|$.