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Doubt about exercise with eigenvalues

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data


    Given the endomorphism ϕ in ##\mathbb{E}^4## such that:
    ϕ(x,y,z,t)=(4x-3z+3t, 4y-3x-3t,-z+t,z-t) find:

    A)ker(ϕ)
    B)Im(ϕ)
    C)eigenvalues and multiplicities
    D)eigenspaces
    E)is ϕ self-adjoint or not? explain

    3. The attempt at a solution

    I get the associated matrix:

    (4 0 -3 3)
    (0 4 -3 -3)
    (0 0 -1 1)
    (0 0 1 -1)

    but i can remove the last row, because it equals the third multiplied by -1

    -solving AX=0 i have ker= L((0,0,1,1),(3, -3, 4, 4))

    - reducing the columns i get Im= L((1, 0, 0),(0,1,0))

    I'm not really sure this results are right, but what I wanted to ask is:
    how do I compute the eigenvalues if the matrix is not square? is the rest of the exercise unsolvable?

    thank you :)
     
  2. jcsd
  3. Apr 30, 2013 #2

    micromass

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    I'm not sure why you think you can do this... If you're solving a system of equation Ax=b, then this is a valid technique. But you're not doing that here.
     
  4. Apr 30, 2013 #3
    Okay, so I'll have to keep alle the rows all through the exercise, right?
    I've taken it again, with the following results:

    ##ker(\phi)=L(0, \frac{3}{2}, 1, 1)##

    ##Im(\phi)=L((-3,-3,-1,-1),(4,0,0,0),(0,4,0,0))##

    eigenvalues and multiplicity: ##T_1=0^1 ; T_2=-2^1; T_3=4^2##

    eigenspaces:

    (6 0 -3 3)
    (0 6 0 0)
    (0 0 -3 3) = T(-2)
    (0 6 -5 1)

    ##V_{-2}= L((2,0,-1,1),(0,1,0,0),(0,0,-1,1),(0,0,5,-1))##

    (0 0 -3 -3)
    (0 0 0 0)
    (3 -3 1 -5) = T(4)
    (0 -6 -4 -4)

    ##V_4=L((3,-3,1,5),(0,3,2,2),(0,0,-1,1))##

    ##V_0=ker##

    which is not selfadjoint because its algebric multiplicity is different from its geometrical multiplicity.

    is it okay now?
    thanks again
     
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