Linear Algebra

Topic: Projective Geometry 343
The following result is more evidence of the ‘niceness’ of the geometry of the
projective plane, compared to the Euclidean case. These two triangles are said
to be in perspective from P because their corresponding vertices are collinear.
O
T1
U1 V1
T2
U2
V2
Desargue’s Theorem is that when the three pairs of corresponding sides — T1U1
and T2U2, T1V1 and T2V2, U1V1 and U2V2 — are extended, they intersect
and further, those three intersection points are collinear.
We will prove this theorem, using projective geometry. (These are drawn as
Euclidean figures because it is the more familiar image. To consider them as
projective figures, we can imagine that, although the line segments shown are
parts of great circles and so are curved, the model has such a large radius
compared to the size of the figures that the sides appear in this sketch to be
straight.)
For this proof, we need a preliminary lemma [Coxeter]: if W , X, Y , Z are
four points in the projective plane (no three of which are collinear) then there
is a basis B for R3 such that
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
0
0
1
RepB( �w) = ⎝0⎠
RepB(�x) = ⎝1⎠
RepB(�y) = ⎝0⎠
RepB(�z) = ⎝1⎠
0
0
1
1
where �w, �x, �y, �z are homogeneous coordinate vectors for the projective points.
The proof is straightforward. Because W, X, Y are not on the same projective
line, any homogeneous coordinate vectors �w0, �x0, �y0 do not line on the same
plane through the origin in R 3 and so form a spanning set for R 3 . Thus any
homogeneous coordinate vector for Z can be written as a combination �z0 =
a · �w0 + b · �x0 + c · �y0. Then �w = a · �w0, �x = b · �x0, �y = c · �y0, and�z = �z0 will do,
for B = 〈 �w, �x, �y〉.
Now, to prove of Desargue’s Theorem, use the lemma to fix homogeneous
coordinate vectors and a basis.
⎛
RepB(�t1) = ⎝ 1
⎞
⎛
0⎠
RepB(�u1) = ⎝
0
0
⎞
⎛
1⎠
RepB(�v1) = ⎝
0
0
⎞
⎛
0⎠
RepB(�o) = ⎝
1
1
⎞
1⎠
1