Linear maps

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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
23
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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
 
  • #3
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Yea that makes sense. Does anyone know about question iii? Thats the part i'm not sure about really
 
  • #4
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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.
 

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