Mathematics 152
Collineations of Affine Planes
Every linear transformation is a collineation, which means that is maps lines into lines. For R2 
the converse is true: every collineation that holds the origin fixed is a linear transformation and can be represented by a matrix.  But for some finite affine planes this is not so.  Here is a counterexample from the medium affine faculty senate.
Choose the athletics committee as x axis, Adam as additive identity 0, Beth as multiplicative identity 1.  Rather arbitrarily, associate the coordinate u with Carl and u+1 with Dave.  Now addition and multiplication can be done by treating the coordinates as polynomials in u, with all arithmetic done modulo 2 and u2 -> u+1.
For multiplication, u2 = u+1, (u+1)2 = u, and u(u+1) = 1.  This agrees with the results obtained geometrically.

Choose the education committee as y axis, Adam as additive identity 0, Emma as multiplicative identity 1.  Rather arbitrarily, associate the coordinate u with Irma and u+1 with Mike.  
Now every instructor has an x and y coordinate.  We can imagine the senate seated like this.

education                 football                      gifts           housing

Mike(0,u+1)              Neil(1,u+1)            Owen(u,u+1)         Phil(u+1, u+1)      diversity

Irma(0,u)                  Jane(1,u)              Kate(u,u)               Lynn(u+1,u)         compensation

Emma(0,1)               Fred(1,1)              Greg(u,1)              Herb(u+1,1)          budget

Adam(0,0)                Beth(1,0)             Carl(u,0)                Dave(u+1,0)         athletics

Every committee now satisfies a linear equation, and it can be obtained by taking a "point" (for example, a member of the education committee) and adding on all multiples of a vector like (1,u).

Consider collineations of the senate that have Adam as a fixed point.  Other instructors move to different seats, but entire committees are mapped to entire committees.  If you select four seats that are occupied by members of a certain committee, the four instructors who move into those seats will all serve on the same committee (alias viewpoint) and the seats into which the original occupants move will all have been occupied by a single committee (alibi viewpoint).

The obvious way to get such a collineation is to perform a linear transformation.  Specify where Beth (1,0) moves -- say to Dave's seat (u+1,0).  Specify where Emma moves -- say to Jane's seat (1,u).  Then the matrix A = acts on the vector for each instructor to determine the new seat.  For example, Kate (u,u) moves to = and takes over Phil's seat.
Lines are mapped into lines, and this is a collineation.  If Adam, Beth, and Emma all stay fixed, the matrix A is the identity, and the resulting collineation is the identity.

But we could initially have associated Dave and Mike with the coordinate u and Carl and Irma with the coordinate u+1. This suggests a collineation of the senate in which the compensation and diversity committees trade places and the gifts and housing committees trade places.  Here is the result.

education                 football                 housing               gifts

Irma(0,u+1)               Jane(1,u+1)            Lynn(u,u+1)        Kate(u+1,u+1)     compensation

Mike(0,u)                 Neil(1,u)                 Phil(u,u)             Owen(u+1, u)      diversity

Emma(0,1)               Fred(1,1)              Herb(u,1)              Greg(u+1,1)          budget

Adam(0,0)                Beth(1,0)             Dave(u,0)                Carl(u+1,0)         athletics

You can check that this is indeed a collineation.  For example, the kitchen committee (Carl, Herb, Irma, and Neil) are now sitting in the seats formerly occupied by the library committee (Dave, Greg, Jane, and Mike).  But Adam, Beth, and Emma kept their seats!  So this is a collineation with three noncollinear fixed points that is not the identity.  There is no way to represent it as a linear transformation.

That means that the theorem "Every collineation is an affine transformation" does not hold for incidence geometry in general.  But it is true in the Euclidean plane. Why?  Because our vector space is over the real numbers, and the real numbers do not have anything like the arbitrary choice of Carl or Dave as "u."  Once the additive and multiplicative identities are assigned on a real line, every other point has an unambiguous coordinate assignment.

To put it more formally, the athletics committee (a field with four elements) has a nontrivial automorphism (interchanging u and u+1) that preserves addition and multiplication.  For the real numbers the only automorphism is the identity. The same is true for the large and small senates, in which choosing "0" and "1" assigns a coordinate unambiguously to every point.

Let f be a collineation of an affine plane) such that
 f(0) = 0, f(e1) = e1, and f(e2) = e2.
Then f is the identity.  Here e1 and e2 are the basis vectors (1,0) and (0,1).
If this is a linear transformation, it is represented by the identity matrix.

For a collineation with f(0) = 0, if v and w are linearly independent vectors, then

f(v + w) = f(v) + f(w)
Note: this is only part of what it takes to make f a linear transformation!

We construct the vector v+w geometrically as the intersection of the line m through v parallel to w with the line n through w parallel to v.  So f(v+w) must be the intersection of f(m) with f(n).

But f(m) passes through f(v) and is parallel to f(w)
f(n) passes through f(w) and is parallel to f(v).
So when we form f(v) + f(w) we also find the intersection of f(m) and f(m) and obtain the same point of intersection.

The x axis, containing 0 and e1, has two fixed points so it is a fixed line.
But we cannot assume that every point on it is fixed.  So assume f(te1) = f(t) e1,
Since 0 is fixed, f(0) = 0.   Since e1 is fixed, f(1) = 1.
Similarly, assume f(te2) = y(t) e2.
(The functions f and y might, for example, interchange Carl and Dave
 or Irma and Mike.)

Now look at the effect of f on the line through e1 + e2.
Since f(e1 + e2) = f(e1) + f(e2) = e1 + e2, this line has two fixed points, the origin and e1+e2, so it must be mapped into itself by f
 f(te1 + te2) = f(te1) + f(te2) = f(t) f(e1) + y(t) f(e2) = f(t) e1 + y(t) e2 
But f(te1 + te2) must be a multiple of e1 + e2, since the line through e1 + e2 is fixed
So f = y : that is why we had to swap both Carl with Dave and Irma with Mike.

Recall that in assigning coordinates for the medium faculty senate, we made the arbitrary
choice of "who is u and who is u+1" just once, on the committee that includes the origin and e1+e2.

Now consider a line of arbitrary slope m through the origin.
f(e1 + me2) = f(e1) + f(m) f(e2) = e1 + f(m) e2.
All multiples of e1 + me2 lie on a line; so their images lie on the line through the origin and
e1 + f(m) e2.

So f(xe1 + xme1) = f(xe1) +  f(xme2) = f(x)e1 + f(xm) e2.
This must be some multiple of e1 + f(m) e2: the multiple is f(x), and f(xm) = f(x)f(m).

Finally, consider f( te1 + s e1) = f(t+s) e1.
 Since we have the sum of two proportional vectors, our result from the preceding page does not apply, so we add and subtract se2.
f(te1 + se2 + s e1 + (-s)e2) = f(te1 + se2) + f(s e1 + (-s)e2) =
f(t) e1 + f(s) e2 + f(s) e1 + f(-s) e2 =
(f(t) + f(s)) e1 (since f(-s) = f(-1) f(s) = -f(s)
Therefore f(t+s) = f(t) + f(s)
Our conclusion is that if f is a collineation that fixes 0, e1, and e2, then
f(0) = 0, f(1) = 1, f(-1) = -1, f(t)f(s) = f(ts), and f(t+s) = f(t) + f(s).

So for an arbitrary vector v = xe1 + ye2
f(tv) = f(txe1 + tye2) = f(tx) e1 + f(ty) e2 = f(t)f(x) e1 + f(t)f(y) e2 = f(t) f(v).
This property characterizes a so-called semilinear transformation. If f is the identity function, then the transformation is linear.

For Zp, a bijection that satisfies
 f(s + t) = f(s) + f(t) and f(st) = f(s) f(t)
must be the identity.
Proof: Set s = 0. f(t) = f(0) + f(t) so f(0) = 0.
Set s = 1.  f(t) = f(1) f(t) so f(1) = 1
(That's why Adam and Beth had to stay fixed)
Set s = 1, t = -1.  f(0) = f(1) + f(-1) so f(-1) = -1.
Set s = 1, t = 2.   f(2) = f(1) + f(1) so f(2) = 2.
Set s = n, t = 1.   f(n+1) = f(n) + f(1) = f(n) + 1.
So by induction f(n) = n for all positive integers n up to p-1.

In the small or large affine senate, any instructor can be reached by adding together the additive identity enough times, so this completes the proof.  But in the medium senate, or in any other finite field whose order is a prime power pn with n > 1, there are elements like u for which this step of the proof does not apply. 

For the real numbers we continue the proof as follows:
Set s = -1, t = n.  f(-n) = f(-1) f(n) = -n
Set q = m/n so that m = nq
Then f(m) = f(n) f(q) or m = n f(q) so f(q) = m/n.
Thus f is the identity on the rational numbers.

For any positive a, a = b2, so f(a) = f(b)2 > 0
f(a) could not be zero, because f is a bijection and f(0) = 0.
So if a > 0, f(a) > 0.

Now choose any t and s with t > s.
Then f(t -s) = f(t) - f(s) > 0 : if t > s, then f(t) > f(s)
So f must preserve ordering.
The final step is to rule out the possibility that f(t) might not equal t for some irrational t.
The proof is by contradiction, and it relies on the fact that there is a rational number between any two real numbers.
Suppose f(t) > t.  There is a rational number q such that
t < q < f(t).
But f(q) = q, so f(q) < f(t) while q > t -- a contradiction.
Assuming f(t) < t leads to the same contradiction. So f(t) = t for any real t.