Mathematics 138
Linear Algebra and Affine Geometry
A collineation is a bijection (1-to-1 onto mapping) T: E2 -> E2 that carries lines into lines: points P, Q, R are collinear if and only if their images TP, TQ, TR are collinear.  It follows that collineations preserve incidence properties.

1. If P and Q determine line m then TP and TQ determine line Tm. 

Proof: Parameterize the line m with the real number t.
A point on m is of the form X(t) = (1-t)P + Q.
Now apply T:
TX(t) = (1-t)TP + t TQ
This shows that the image of any point on m is on the line determined by TP and TQ.

2. If lines m and n intersect in point R, Tm and Tn intersect in TR.
Proof: Point R is both on line m and line n.
So point TR is both on Tm and Tn and is their (unique) point of intersection.

3. If lines m and n are parallel, so are Tm and Tn.
Proof: Since T is a bijection, it is invertible.
Do a proof by contradiction.

Suppose that Tm and Tn are not parallel.  Then they intersect in some point S.
Apply T-1: by the preceding result, m and n intersect in T-1 S and hence are not parallel.
Conclusion: Tm and Tn are parallel.
An affine transformation T: E2 -> E2  has the form Tx = Ax + b, where A is any invertible 2x2 matrix and b any vector in R2.  
Generally true, and easy to prove:
Every affine transformation is a collineation.
Three things to show:
1. T is surjective (onto)
Every point y must be in the image of T.

Ax + b = y.
Ax = y - b.
x = A-1 (y - b) (good thing that A is invertible)

2. T is injective (1-to-1)

Assume the contrary: Ax + b = A x' + b for distinct x and x'.
Then A(x - x') = 0, which contradicts the assumption that A is invertible.

3. The image of a line is another line.
The line m determined by P and Q is

P + t (Q-P)  or P + t v  or   (1-t) P + t Q   where v = Q - P

Its image is

Tm = A(P + tv) + b = AP + t Av + b = AP + b + t A v = TP + t (Av) = TP + t (AQ - AP)

which is also a line.

Note: if both P and Q are fixed, then

Tm = TP + t (TQ - TP) = P + t (Q - P)
so every point on m is fixed.
True for the real affine plane R2, but not in general, and harder to prove:
Any collineation is an affine transformation. (Ryan, Appendix E).
We'll return to this.
The fundamental theorem of affine geometry
Key lemma: Ryan Ch. 2, Theorem 7
If an affine transformation leaves fixed three noncollinear points, it must be the identity. 

Given any X (not necessarily inside the triangle PQR) construct a line through X that intersects any two sides of the triangle (possible by "Pasch's Axiom)."
If P and Q are fixed, so is B.
If Q and R are fixed, so is A.
If A and B are fixed, so is X.

The fundamental theorem:
Given three noncollinear points P, Q, R and their images P', Q', and R', there is a unique affine transformation T such that TP = P',  TQ = Q', and TR = R'.
Proof: Here is the recipe for constructing the transformation. 

Let B be the matrix that carries v =   into Q-P and w =  into R - P
Let C be the matrix that carries v =  into Q'-P' and w =  into R' - P'
The matrix CB-1 carries x - P into x' - P' 
Let b be the vector P' - P
Then T x = CB-1 (x-P) + P' 
If x = P, T x = 0 + P'
If x = Q, T x = CB-1 (Q-P) + P' = Q' - P' + P' = Q'
If x = R, T x = CB-1 (R-P) + P' = R' - P' + P' = R'

Uniqueness: Suppose there were two transformations S and T that do this.
Then S-1T fixes P, Q and R.  So, by the lemma above, it is the identity
and S = T.

Important examples of affine transformations that are not Euclidean isometries:

With determinant +1
Shears (carry rectangles into parallelograms)

With determinant -1
"Affine reflections" (eigenvalues +1 and -1, but the eigenvectors are not orthogonal)

Both of these do not preserve angles; so they are not particularly important for Euclidean geometry.

With determinant not equal to 1:

A central dilatation moves each point along the line joining it to the center C, multiplying the
vector x - C by a magnification factor k that is different from 1.

So T x = C + k (x - C) = k x + ( 1 - k ) C

Similarities preserve angles but multiply all distances by a magnification k > 0:

d(Tx, Ty) = k d(x, y) 

A similarity can always be realized as a dilatation followed by an isometry.  First a geometric figure is enlarged or reduced, then it is moved into its final position. 

Proof: Let S be the central dilatation with magnitude 1/k whose center is the origin, so that
d(Sx, Sy) = (1/k) d(x,y) .

Then the product TS has magnification 1 and is an isometry.

So T = (TS) S-1 is a central dilatation (whose center is the origin) followed by an isometry.

The 3x3 matrix for a similarity is like that for an isometry, except that the first and second columns have magnitude k instead of 1.  The collection of all similarities with nonzero magnification is a group.  If T1 has magnification k1 and T2 has magnification k2 then T1 T2 has magnification k1 k2 and the inverse of T1 has magnification 1/k1.

Similarities are important because of a peculiar property of Euclidean geometry: the existence of similar triangles, which have congruent angles but whose sides are not equal, merely proportional. In other geometries like elliptic and hyperbolic, triangles are similar only if they are congruent.

In developing the Euclidean theory of similar triangles, we can say that two figures are similar if there is a similarity that transforms one into the other.

Now back to the deferred issue: is every collineation an affine transformation?  Here is a frightening 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.

So the key step in the proof of Ryan's Theorem 2, which is "too technical to present here but is included is Appendix E" is to show that the real number system has no nontrivial automorphisms. In studying this proof, is is important to distinguish between what is true for arbitrary affine geometry and what is true only for affine geometry over certain fields where choosing "0" and "1" assigns a coordinate unambiguously to every point.

The theorem proved in Appendix E:
Let f be a collineation (of the Euclidean 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).
Lemma 1(true for any affine geometry)
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 the head of v and parallel to w, with the line n, through the head of w and 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.
Part of the proof of the theorem (true for any affine geometry)
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 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).
Setting x = m = -1, we see also that f(1) = f(-1) f(-1) so f(-1) = -1.

Finally, consider f( te1 + s e1) = f(t+s) e1.
 Since we have the sum of two proportional vectors, Lemma 1 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.

Lemma 2: For the real numbers, 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 this point the proof holds for the small and large affine senates, where any instructor can be reached by adding on the additive identity enough times.)
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.

The final step is to extend the proof to the real numbers.  To do this we will rely on properties of the real numbers that do not hold for finite fields or for complex numbers, namely

The real numbers are an ordered field: there are positive elements whose sums and products are positive, and every positive element has a positive square root.

We can easily show that f carries positive elements into positive elements.
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

Conclusions:
A collineation of the Euclidean plane or the small or large faculty senate that fixes three noncollinear points must be the identity.

For this case (though not for the medium faculty senate)
Suppose an affine transformation T maps P to P', Q to Q', R to R' 
Suppose some other collineation S maps P to P', Q to Q', R to R' 
Then T-1 S fixes P, Q, and R and must be the identity.
So S = T and any collineation is an affine transformation.

For the medium faculty senate there exist collineations that are not affine transformations.
The proof above fails because T-1 S is not necessarily the identity.
It could be the collineation which interchanges the two other members of the committee with P and Q while also interchanging the two other members of the committee with P and R.