5 points determine uniquely a conic 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.
 A: "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.


*

*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.

*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.

*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 $ax(x-1) + by(y-1) = 0$.
