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

Homework Help: Linear algebra - Spectral decompositions: Eigenvectors of projections

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Let P1 and P2 be the projections defined on R^3 by:

    P1(x1, x2, x3) = (1/2(x1+x3), x2, 1/2(x1+x3))
    P2(x1, x2, x3) = (1/2(x1-x3), 0, 1/2(-x1+x3))

    a) Let T = 5P1 - 2P2 and determine if T is diagonalizable.
    b) State the eigenvalues and associated eigenvectors of T.


    2. Relevant equations



    3. The attempt at a solution


    For a), I believe it is diagonalizable because P1 + P2 gives us (x1, x2, x3). Although I'm could be wrong on that...

    It's mainly b) that I'm concerned for. By the theorem, (T = c1P1 + c2P2...+.. where c are eigenvalues) 5 and -2 are the eigenvalues (although this sort of confuses me because I had thought the eigenvalues of projections are always 1 and 0).

    How can we find the eigenvectors? Had this been a matrix it's simple, subtract the eigenvalue from the main diagonal, simplifiy, and find the nullspace.



    Also, the solution to b) is for eigenvalue 5, the eigenvectors are <(1,0,1),(0,1,0)> and for eigenvalue -2, the eigenvector is <(-1,0,1)> (so my guess some matrix is formed?)


    It feels like I'm missing something obvious here. Can anyone please help me out?

    Thanks
     
  2. jcsd
  3. Apr 19, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I don't see how that follows. Is there some theorem that you're using?
    T isn't a projection, so its eigenvalues don't have to be 0 or 1.
    You should be able to write down the matrices for P1 and P2 by inspection, and then you can calculate the matrix for T. Or you can find the nth-column of the matrix for T by applying T to the nth basis vector.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook