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I am studying a linear method for triangulation that uses Singular Value Decomposition (SVD) to estimate a 3D point P from multiple camera views. The method formulates a linear system of equations $AP=0$ and finds the best least-squares solution by taking the last column of the SVD result for matrix A, which corresponds to the solution vector P with the constraint that $P$ .

However, the textbook I'm reading states that this method is "not suitable for projective reconstruction" because it's not "projective-invariant." The text gives the following explanation:

If we replace the camera matrices $M, M'$ with ones affected by a projective transformation $H$, say $M H^{-1}, M'H^{-1}$, a solution P to the previous estimation of $AP=0$ will correspond to a solution $HP$ for the transformed problem. Recall that SVD solves for the constraint $|P|=1$, which is not invariant under a projective transformation H. Therefore, this method... is often not the optimal solution to the triangulation problem.

I'm struggling to understand the exact reason why the SVD's normalization constraint breaks the projective invariance. I have a few specific questions:

  1. What is the precise reason the SVD's normalization ($|P|=1$) prevents this method from being projective-invariant? In homogeneous coordinates, two vectors are equivalent if they are scalar multiples of each other. Since SVD forces the solution to be a unit vector, how does this process "break" the underlying projective relationship?

  2. The solution P is a homogeneous coordinate vector. We can get the Euclidean coordinates of the 3D point by simply dividing by the last component (the $w$ value). If the SVD solution in the transformed system is Q, and we know that $Q \sim HP\_{real}$ (where $P\_{real}$ is the real point), why can't we just find the Euclidean coordinates from Q? Does this not give a valid result that is at least projectively equivalent to the original point?

  3. The text says the solution becomes $HP$. If we can get a new solution Q from the transformed problem, why can't we just compute $P = H^{-1}Q$ and then normalize it? Does this not give a valid result for P? My intuition is that even if the scale factor $\\lambda_H$ changes, the underlying homogeneous coordinate relationship should still hold.

I would appreciate a clear, mathematically-grounded explanation of this concept. Thank you for your time and expertise.

https://web.stanford.edu/class/cs231a/course_notes/Course_Notes_4.pdf This is source and my confusion lies in the last paragraph of section 2.1: A nonlinear method for triangulation

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If the system of linear equations $AP = 0$ has a non trivial solution $P \ne 0$, then it has an infinite number of such solutions. In particular, if $P=x_0$ is a solution, then $P=\lambda x_0$ is also a solution for any real $\lambda$ because $A \lambda x_0=\lambda A x_0=\lambda\cdot 0=0$.

If a non trivial solution exists, the SVD method chooses the right singular vector (of unit length) corresponding to a singular value of zero. If no non trivial solution exists, the SVD method does the next best thing: it chooses the right singular vector (of unit length) corresponding to the smallest singular value. This choice of unit length is the constraint $||P||=1$ that is mentioned in the notes.

Projective transformations do not preserve the length of vectors and so the constraint $||P||=1$ becomes meaningless. You are right that even after a projective transformation the solution still satisfies $AP = 0$ (assuming there was a singular value of zero to begin with), but it is unlikely to satisfy $||P||=1$.

Though it is not explicitly mentioned in the notes, the singular values themselves are not invariant under a general projective transformation and the choice of the smallest singular value also does not make sense anymore.

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