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Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 258 to 259. In particularenter image description here 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 $\| t\|_2=1$?

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2 Answers 2

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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 $2(t_x^2+t_y^2+t_z^2) = 2\|\mathbf t\|$.

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  • $\begingroup$ Sorry for my poor math background, but how to understand "restricted to the complement of its null space"? whether or not it means spanning the rowspace of $E$ by basis $\{v_1,v_2\}$, because complement of its nullspace is its rowspace and $Ev_3=Udiag(1,1,0)V^Tv_3=0$. If not, what does the "restrict" mean? $\endgroup$
    – Finley
    Dec 27, 2017 at 12:38
  • $\begingroup$ @Finley I could’ve said row space, too, but was thinking more of $\mathtt E$ as a linear map than as a matrix at the time. The definition of a function includes its domain; the restriction of a function is a derived function whose domain is a subset of the original one. In set-theoretic terms, it’s a subset of the function. $\mathtt E$ as a whole isn’t length-preserving since it “collapses” a certain direction, but if you look only at its action on its row space, the singular values are all $1$ (eigenvalues $\pm1$), so that restricted map is an isometry. $\endgroup$
    – amd
    Dec 27, 2017 at 18:00
  • $\begingroup$ Thanks to your explanation of "restricted", now I can derive $E(\alpha_1v_1+\alpha_2v_2)=Udiag(0,0,1)V^T(\alpha_1v_1+\alpha_2v_2)=(\alpha_1u_1+\alpha_2u_2)$ , so $E$ is surely a isometry when restricted to it's complement of null space(i.e. row space). But how to employ this fact to show $S$ is also a isometry? $\endgroup$
    – Finley
    Dec 28, 2017 at 5:13
  • $\begingroup$ $S(\alpha_1v_1+\alpha_2v_2)=ER^T(\alpha_1v_1+\alpha_2v_2)$, the inner map $R^T$ is length-perserving but $R^T$ can map the vector outside $E$'s row space, so the outer map $E$ will not be restricted to it's row space. $\endgroup$
    – Finley
    Dec 28, 2017 at 5:19
  • $\begingroup$ @Finley $S$ is not an isometry, nor could it be: it’s rank-deficient. I’ve clarified in my answer that it’s the restriction of $S$ to its row space that’s length-preserving. I don’t understand what point you’re trying to make about $ER^T$. $E$ and $S$ don’t have the same row spaces, but so what? $\endgroup$
    – amd
    Dec 28, 2017 at 5:41
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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.

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  • $\begingroup$ Nicely done. Note that there might be other square roots of $S^2$ as well. $\endgroup$
    – amd
    Dec 29, 2017 at 19:47
  • $\begingroup$ Yeah, I cannot derive the factorization uniqueness, but I think the factorization feasibility proved above is enough. Thanks for your patience! :) $\endgroup$
    – Finley
    Dec 30, 2017 at 3:22
  • $\begingroup$ If you go back a couple of pages in the book, you’ll see that the factorization of $E$ isn’t unique because of the repeated singular value. Similarly, $S$ is a rotation of its column space, so you can use any orthonormal basis of that space in $U$ in its decomposition. $\endgroup$
    – amd
    Dec 30, 2017 at 7:07

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