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I'm reading "Multiple View Geometry" by Richard Hartley and Andrew Zisserman. On pages 30-31 they have a section which proves that 5 points determine a conic. now, I know this question has been asked before but I would like an answer for the specific presentation which I describe below. First, they describe a conic by the equation: $$ax^2+bxy+cy^2+dx+ey+f=0$$ So by aggregating 5 points $(x_i,y_i)$ for $i\in\{1,\ldots,5\}$ which hold the above equation we get the matrix equation: $$\begin{bmatrix}x_1^2&x_1y_1&y_1^2&x_1&y_1&1\\x_2^2&x_2y_2&y_2^2&x_2&y_2&1\\x_3^2&x_3y_3&y_3^2&x_3&y_3&1\\x_4^2&x_4y_4&y_4^2&x_4&y_4&1\\x_5^2&x_5y_5&y_5^2&x_5&y_5&1\end{bmatrix}\mathbf{c}=0$$ where $\mathbf{c}=(a,b,c,d,e,f)^T$. Finally, they claim the following:

"...the conic is the null vector of this $5\times$6 matrix . This shows that a conic is determined uniquely (up to scale) by five points in general position."

I have 2 questions about their derivation:

1) I know from other sources that 5 points where no 3 of them are collinear determine uniquely a conic so I guess that here the authors meant general linear position (which in this case translate to the condition that no 3 of the 5 points are collinear) and not some other kind of notion of general position. Am I right?

2) From a Linear algebraic point of view I would say that the immediate conclusion of the above matrix equation is that if the rank of such matrix is 5 then those 5 points determine uniquely a conic so I guess that this condition (of rank=5) can be translated to the condition for the 5 points to be in general position but I can't see why it is the same.

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  • $\begingroup$ Keep reading. “General position” is defined a few pages later. $\endgroup$ – amd Nov 14 '19 at 17:49
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"General position" in geometry means all points on which some polynomial (or finite set of polynomials) is non-zero. For example, "a general matrix is invertible" is true because a matrix is invertible if its determinant is non-zero. So to say that 5 points in general position simply means that if that matrix has rank 5 (meaning the 5x5 minors are non-zero), then they determine a unique conic.

The harder question is then to classify which points give you a matrix with rank < 5. If you look at the equations, you have 10 variables and 5 equations. Most likely the author's did not intend any further analysis beyond: if this matrix has rank 5 then there is a unique conic. Also, to analyze this yourself, you want to be more clever than just looking at the minors of that matrix.

Here's what Wikipedia says about the same argument:

The two subtleties in the above analysis are that the resulting point is a quadratic equation (not a linear equation), and that the constraints are independent. The first is simple: if A, B, and C all vanish, then the equation $D x + E y + F = 0$ defines a line, and any 3 points on this (indeed any number of points) lie on a line – thus general linear position ensures a conic. The second, that the constraints are independent, is significantly subtler: it corresponds to the fact that given five points in general linear position in the plane, their images in $\mathbf{P}^5$ under the Veronese map are in general linear position, which is true because the Veronese map is biregular: i.e., if the image of five points satisfy a relation, then the relation can be pulled back and the original points must also satisfy a relation. The Veronese map has coordinates $[x^{2}:xy:y^{2}:xz:yz:z^{2}]$, and the target ${\mathbf {P}}^{5}$ is dual to the $[A:B:C:D:E:F]$ ${\mathbf {P}}^{5}$ of conics. The Veronese map corresponds to "evaluation of a conic at a point", and the statement about independence of constraints is exactly a geometric statement about this map.

The Veronese map that they are referring to is $f : \mathbf{P}^2 \to \mathbf{P}^5$ given by $$ f([x:y:z]) = [x^2 : xy : y^2 : xz : yz : z^2] $$

It is easier to homogenize (i.e. to replace $x,y,1$ by $xz, yz, z^2$) then the original equation is when $z = 1$.

What they say now is that $f$ is "biregular" meaning invertible by polynomial expressions. In particular, this means that given $[x^2 : xy : y^2 : xz : yz : z^2]$, you can determine $x, y, z$. To analyze exactly what happens to linear relations under $f$, we should know how to invert it. Basically you set $z = 1$ and then pick out the $xz$ and $yz$ coordinates.

So now let's say that the matrix has rank $4$ or less. Then there is some relation

$$ \lambda_1 f([x_1:y_1:z_1]) + \dots + \lambda_5 f([x_5:y_5:z_5]) = 0 $$

By inverting $f$, this gives

$$ \lambda_1 [x_1:y_1:z_1] + \dots + \lambda_5 [x_5:y_5:z_5] = 0. $$

Note that this doesn't work the other way. However, what this does say is that all relations between $f([x_i:y_i:z_i])$ come (in some way) from linear relations between $[x_i:y_i:z_i]$. So all we need to consider now is linear relations between points and what those relations say about uniqueness.

  1. If three points are collinear but the other two do not lie on this line then there is still a unique (degenerate) conic: the unique line passing through the three collinear points + the unique line passing through the other two.

  2. If four points are collinear then there is not a unique conic since you can (and must) take a line passing through those 4 points and then there is not a unique line passing through the other point.

  3. If five points are collinear it is a similar picture as 2. except the second line doesn't even need to pass through a point anymore.

You could also have some points being equal. For example if you have 4 points then you can find several pairs of lines that go through all four points, as well as some nondegenerate conics. For instance with the points $(0,0), (0,1), (1,0), (1,1)$. The two pairs of lines $x(x-1) = 0$ and $y(y-1) = 0$ pass through all four points. The set of all conics that pass through these points is $ax(x-1) + by(y-1) = 0$.

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