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I'm having a hard time understanding how a few equations are being derived. So the fundamental equation is an equation that relates corresponding points in stereo images. Anyway, that's the basic background. The question is more on the matrix manipulation. So I'm trying to get (1) and (2) into (3) and I have

(1) $\tilde m_1 = A_1[I\hspace{5 pt}0]\tilde M $

(2) $\tilde m_2 = A_2[R\hspace{5 pt}t]\tilde M$

Eliminating $M$ from the above equations you can obtain

(3) $\tilde m_2^T \hspace{2 pt} A_2^{-T} \hspace{2 pt} [t]_x\hspace{2 pt} R \hspace{2 pt}A_1^{-1} \tilde m_1 = 0$

And some notes:

a. If $V=[x,y,..]^T$, then $\tilde V = [x,y,..,1]^T$

b. $[t]_x$ is an antisymmetrix (or skew symmetric) matrix such that $[t]_xx = t \times x$ for all vector x

c. $[A\hspace{5 pt}b]$ is a matrix composed of $A$ with vector $b$ as the last column

d. $\tilde m$ is 3x1, $\tilde M$ is 4x1, $A$ is 3x3, and $R$ is 3x3

So far I can get as far as dropping the scalars, then taking $A^{-1}$ of both sides of (1) but beyond that I'm a bit lost...Perhaps if I could get some hints into matrix operations that would be useful or identities that can simplify some of these

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This website is for questions of mathematical research interest. I'm not convinced there is any research interest in this question. It might fit better on math.stackexchange.com –  Gerry Myerson Mar 24 '13 at 3:58
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1 Answer

up vote 0 down vote accepted




into the left-hand side of equation (3),


because $x^T Ax=0$ for any antisymmetric matrix $A$.


it is perhaps more clear if I write the last line out into components, abbreviating $A=[t]_x$ and $v=RM$:


$\;\;=\sum_{n,m=1}^{3}(v_n+t_{n})A_{nm}v_{m}=\sum_{n,m=1}^{3}t_{n}A_{nm}v_{m}=-v\cdot(t\times t)=0$

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Okay, I can understand that but what I'm not understanding is how you can say M^T[R t]^T is the transpose of R[I 0]M. How do you deal with the t? –  Jared Joke Mar 23 '13 at 22:06
I wrote the whole thing out into components. Is it clear now? –  Carlo Beenakker Mar 24 '13 at 11:39
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