Why does fundamental matrix have 7 degrees of freedom?

Question

There are 9 parameters in the fundamental matrix to relate the pixel co-ordinates of left and right images but only 7 degrees of freedom (DOF).

The reasoning for this on several pages that I've searched says :

1. Homogenous equations means we lose a degree of freedom

2. The determinant of F = 0, therefore we lose another degree of freedom.

I don't understand why those 2 reasons mean we lose 2 DOF - can someone explain it?

Answer 1

We initially have 9 DOF because the fundamental matrix is composed of 9 parameters, which implies that we need 9 corresponding points to compute the fundamental matrix (F). But because of the following two reasons, we only need 7 corresponding points.

Reason 1

We lose 1 DOF because we are using homogeneous coordinates. This basically is a way to represent nD points as a vector form by adding an extra dimension. ie) A 2D point (0,2) can be represented as [0,2,1], in general [x,y,1]. There are useful properties when using homogeneous coordinates with 2D/3D transformation, but I'm going to assume you know that.

Now given the expression p and p' representing pixel coordinates:

p'=[u',v',1] and p=[u,v,1]

the fundamental matrix:

F = [f1,f2,f3]
    [f4,f5,f6]
    [f7,f8,f9]

and fundamental matrix equation:

(transposed p')Fp = 0

when we multiple this expression in algebra form, we get the following:

uu'f1 + vu'f2 + u'f3 + uv'f4 + vv'f5 + v'f6 + uf7 + vf8 + f9 = 0. 

In a homogeneous system of linear equation form Af=0 (basically the factorization of the above formula), we get two components A and f.

A:

[uu',vu',u', uv',vv',v',u,v,1] 

f (f is essentially the fundamental matrix in vector form):

[f1,f2'f3,f4,f5,f6,f7,f8,f9]

Now if we look at the components of vector A, we have 8 unknowns, but one known value 1 because of homogeneous coordinates, and therefore we only need 8 equations now.

Reason 2

det F = 0.

A determinant is a value that can be obtained from a square matrix.

I'm not entirely sure about the mathematical details of this property but I can still infer the basic idea, and, hopefully, you can as well.

Basically given some matrix A

A = [a,b,c]
    [d,e,f]
    [g,h,i]

The determinant can be computed using this formula:

det A = aei+bfg+cdh-ceg-bdi-afh

If we look at the determinant using the fundamental matrix, the algebra would look something like this:

F = [f1,f2,f3]
    [f4,f5,f6]
    [f7,f8,f9]

det F = (f1*f5*f8)+(f2*f6*f7)+(f3*f4*f8)-(f3*f5*f7)-(f2*f4*f9)-(f1*f6*f8)

Now we know the determinant of the fundamental matrix is zero:

det F = (f1*f5*f8)+(f2*f6*f7)+(f3*f4*f8)-(f3*f5*f7)-(f2*f4*f9)-(f1*f6*f8) = 0

So, if we work out only 7 of the 9 parameters of the fundamental matrix, we can work out the last parameter using the above determinant equation.

Therefore the fundamental matrix has 7DOF.

Answer 2

The reasons why F has only 7 degrees of freedom are

1. F is a 3x3 homogeneous matrix. Homogeneous means there is a scale ambiguity in the matrix, so the scale doesn't matter (as shown in @Curator Corpus 's example). This drops one degree of freedom.
2. F is a matrix with rank 2. It is not a full rank matrix, so it is singular and its determinant is zero (Proof here). The reason why F is a matrix with rank 2 is that it is mapping a 2D plane (image1) to all the lines (in image 2) that pass through the epipole (of image 2).

Hope it helps.