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



Deriving the fundamental equation (with regards to computer vision)
substitute


$A\_1^{-1}\tilde{m}\_1=[I\;0]\tilde{M}$


$A\_2^{-1}\tilde{m}\_2=[R\;t]\tilde{M}$


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


$\tilde{M}^{T}[R\;t]^{T}[t]\_{x}R[I\;0]\tilde{M}=(RM)^T[t]\_x(RM)=0$


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


-----update-----


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


$\tilde{M}^{T}[R\;t]^{T}AR[I\;0]\tilde{M}=\left([R\;t]\tilde{M}\right)^{T}A\left([R\;0]\tilde{M}\right)=$


$\;\;=\sum\_{n,m=1}^{3}(v\_n+t\_{n})A\_{nm}v\_{m}=\sum\_{n,m=1}^{3}t\_{n}A\_{nm}v\_{m}=-v\cdot(t\times t)=0$


