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This is from Hartley-Zisserman's Multiple View Geometry in Computer Vision pg 55-56.

We are given a conic $$C_\infty^* = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 0 \end{bmatrix}$$ and we know that there is some unknown projectivity $H \in PGL(3, \mathbb{R})$ such that $C_\infty^{'*} = HC_\infty^*H^T$. The book goes on to claim that we can write the SVD of $C_\infty^{'*}$ as $$C_\infty^{'*} = U \begin{bmatrix} 1 & 0 & 0 \\ 0& 1 & 0 \\ 0& 0 & 0 \end{bmatrix} U^T$$ and then $U = H$ up to a similarity transformation.

This is the part I don't understand. The decomposition above is just a orthogonal diagonalization of a symmetric matrix $C_\infty^{'*}$ but there reason apriori that $C_\infty^{'*}$ has the same eigenvalues $1,1,0$ as $C_\infty^{'*}$ is obtained from $C_\infty^*$ by a congruence rather than a similarity relation. For example if $H = \verb|diag|(1,2,3)$, we have $HC_\infty H^T= \verb|diag|(1,4,0)$. Can you help me see why this is true, and if it is not true, help me see what the author meant here?

Also a side comment is that the Hartley/Zisserman book states many results without proof or comment or citation, and I find it quite difficult each time I stumble on something. I would say that I have a reasonable background in Linear algebra, so I don't think the issue is that I'm not prepared for this book. Any advice on how to tackle this issue would be appreciated. Is there another book which has all these details?

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  • $\begingroup$ I advise you the first part of the excellent book "Computational line geometry" by Pottmann and Wallner (Springer). $\endgroup$
    – Jean Marie
    Aug 29, 2022 at 9:25
  • $\begingroup$ In fact $U^T=U^{-1}$ because $U$ is orthogonal ($U^TU=I_3$). $\endgroup$
    – Jean Marie
    Aug 29, 2022 at 9:27
  • $\begingroup$ Thank you! I will look at that book. And yes, that is true. I mentioned above that it is an orthogonal diagonalization. $\endgroup$
    – matpiliya
    Aug 29, 2022 at 9:39
  • $\begingroup$ ... Therefore, as $UAU^T$ has become $UAU^{-1}$ this matrix has the same eigenvalues as $A$ ! $\endgroup$
    – Jean Marie
    Aug 29, 2022 at 9:44
  • $\begingroup$ That is true, but I think you may have misunderstood the question. We start with A = diag(1,1,0) and then calculate HAH^T for an invertible H. Then it is claimed that you can diagonalize that matrix orthogonally in the form UAU^T. I think that is not correct $\endgroup$
    – matpiliya
    Aug 29, 2022 at 10:09

1 Answer 1

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Take $C^*_\infty$ and ${C^*_\infty}^\prime$ the dual conics of the points $x$ and $x^\prime$ related by the projection $H$. The two conics have the same eigenvalues because $C^*_\infty$ is a degenerate conic therefore it has rank 2 and has a repeated point (pg 32). Moreover, $l_\infty$ is in the null space of $C^*_\infty$ and we have

$$ {C^*_\infty}^\prime {l_\infty}^\prime = HC^*_\infty H^t l_\infty = HC_\infty^* l_\infty = 0 $$

Therefore $l_\infty$ is also in the null space of ${C_\infty^*}^\prime$. If $p$ is the repeated point of $C^*_\infty$ (with eigenvalue 1) we have $C^*_\infty p = p$ and $p^\prime = H^{-T}p = H p$ and we can write

$$ {C^*_\infty}^\prime {p}^\prime = HC^*_\infty H^t H^{-T}p = HC_\infty^* p = Hp = p^\prime $$

So $p$ is also a repeated point of $C^*_\infty$.

About your question about a book having all these details, I think you cant take a look at two books by Olivier Fagueras that are related to the subject of Hartley and Zisserman book. These are:

  1. Three-dimensional computer vision a geometric viewpoint - Olivier Faugeras.
  2. The Geometry of Multiple Images - Olivier Faugeras, Quang-Tuan Luong.
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  • $\begingroup$ Thanks for the references! I will take a look at them. But I'm having a bit of trouble following your answer. Could you kindly elaborate? First, if I understand correctly, $C_\infty^*$ is the dual conic corresponding to the two points $I = [1,i,0]^T$ and $J = [1,-i,0]^T$ so that $C_\infty^* = IJ^T + JI^T$. I believe it would have a repeated point only if it is rank 1. Also, the image of $\ell_\infty$ is $H^{-T} \ell_\infty$, so the equality would be $C_\infty^{*'}\ell_\infty' = HC_\infty^*H^T(H^{-T}\ell_\infty) = 0$. $\endgroup$
    – matpiliya
    Aug 29, 2022 at 10:24
  • $\begingroup$ I feel it cannot be a true result since if $H = \verb|diag|(1,2,3)$, then it will make $C_\infty^{*'} = \verb|diag|(1,2,0)$ which has different eigenvalues $\endgroup$
    – matpiliya
    Aug 29, 2022 at 10:29
  • $\begingroup$ @matpiliya I think what you have misunderstood is that H=U but this is true up to similarity, that is U=H H_s where H_s is a similarity transformation. $\endgroup$
    – Cofinite
    Aug 29, 2022 at 10:57
  • $\begingroup$ Yes, that is true about $H$, but that shouldn't affect $C_\infty^{*'}$ which is unaffected by similarities. The image of the absolute conic under $H = \verb|diag|(1,2,1)$ is $\verb|diag|(1,4,0)$ which clearly cannot be diagonalized with eigenvalues $1,1,0$. $\endgroup$
    – matpiliya
    Aug 29, 2022 at 11:47

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