Linear Algebra/Topic: Projective Geometry/Solutions
- Problem 1
What is the equation of this point?
From the dot product
we get that the equation is .
- Problem 2
- Find the line incident on these points in the projective plane.
- Find the point incident on both of these projective lines.
- This determinant
- Problem 3
Find the formula for the line incident on two projective points. Find the formula for the point incident on two projective lines.
The line incident on
can be found from this determinant equation.
The equation for the point incident on two lines is the same.
- Problem 4
Prove that the definition of incidence is independent of the choice of the representatives of and . That is, if , , , and , , are two triples of homogeneous coordinates for , and , , , and , , are two triples of homogeneous coordinates for , prove that if and only if .
If , , , and , , are two triples of homogeneous coordinates for then the two column vectors are in proportion, that is, lie on the same line through the origin. Similarly, the two row vectors are in proportion.
Then multiplying gives the answer .
- Problem 5
Give a drawing to show that central projection does not preserve circles, that a circle may project to an ellipse. Can a (non-circular) ellipse project to a circle?
The picture of the solar eclipse — unless the image plane is exactly perpendicular to the line from the sun through the pinhole — shows the circle of the sun projecting to an image that is an ellipse. (Another example is that in many pictures in this Topic, the circle that is the sphere's equator is drawn as an ellipse, that is, is seen by a viewer of the drawing as an ellipse.)
The solar eclipse picture also shows the converse. If we picture the projection as going from left to right through the pinhole then the ellipse projects through to a circle .
- Problem 6
Give the formula for the correspondence between the non-equatorial part of the antipodal modal of the projective plane, and the plane .
A spot on the unit sphere
is non-equatorial if and only if . In that case it corresponds to this point on the plane
since that is intersection of the line containing the vector and the plane.
- Problem 7
(Pappus's Theorem) Assume that , , and are collinear and that , , and are collinear. Consider these three points: (i) the intersection of the lines and , (ii) the intersection of the lines and , and (iii) the intersection of and .
- Draw a (Euclidean) picture.
- Apply the lemma used in Desargue's Theorem to get simple homogeneous coordinate vectors for the 's and .
- Find the resulting homogeneous coordinate vectors for 's (these must each involve a parameter as, e.g., could be anywhere on the line).
- Find the resulting homogeneous coordinate vectors for . (Hint: it involves two parameters.)
- Find the resulting homogeneous coordinate vectors for . (It also involves two parameters.)
- Show that the product of the three parameters is .
- Verify that is on the line..
- Other pictures are possible, but this is one.
The intersections , , and are labeled so that on each line is a , a , and a .
- The lemma used in Desargue's Theorem gives a basis with respect to which the points have these homogeneous coordinate vectors.
- First, any on
- Because is we have this.
- Since is the intersection
- Because is on the line its homogeneous coordinate vector has the form
- The homogeneous coordinate vector of can be written in this way.