# Linear maps

mathshelp
Regard the n-dimensional real projective space RPn as the
space of lines in Rn+1 through {0}, i.e.

RPn = (Rn+1 − {0}) /~ with x ~ y if y = λx for λ not equal to 0 ∈ R ;with the equivalence class of x denoted by [x].

(i) Work out the necessary and sufficient condition on a linear map
f : Rn+1 → R m+1 for the formula [f][x] = [f(x)] to define a map
[f]: RPn → RPm ; [x] → [f(x)] :

(ii) For a linear map f : Rn+1 → Rn+1 satisfying the condition of (i)
prove that the fixed point set
Fix([f]) = {[x] ∈ RPn | [x] = [f(x)] ∈ RPn}
consists of the equivalence classes of the lines in Rn+1 through {0} which contain eigenvectors of f.

(iii) Construct examples of linear maps f : R3 → R3 satisfying the condition of (i) such that

(a) Fix([f]) is a point.
(b) Fix([f]) is the disjoint union of a point and a circle.
(c) Fix([f]) is a projective plane.

Mikemaths
for part (ii)
if [x]=[f(x)] in RPn
then by the definition
λx=f(x) so that they are in the same equivalence class therefore since f is linear x is an eigenvector and λ is the eigenvalue

mathshelp
Yea that makes sense. Does anyone know about question iii? Thats the part I'm not sure about really

mrbohn1
Here are some suggestions off the top of my head (read: check the details as they might be wrong!):

(a) Any map which rotates R3 around the origin: then the only fixed point is the origin.

(b) Project R3 onto R2 via the map f:(x,y,z)-->(x,y,0). Then compose this with the map g which fixes zero and takes every other point x-->x/|x|.

This will fix the unit circle: {(x,y): x2+y2=1} and the origin.

(c) Any map f:x-->ax where a is a scalar should fix RP2.