Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear maps

  1. Mar 17, 2010 #1
    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.
  2. jcsd
  3. Mar 17, 2010 #2
    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
  4. Mar 18, 2010 #3
    Yea that makes sense. Does anyone know about question iii? Thats the part i'm not sure about really
  5. Mar 18, 2010 #4
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook