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Linear algebra proof with operator T^2 = cT

  1. Jun 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Let T:V--->V be an operator satisfying T^2=cT c=/=0.
    Show that V=U[tex]\oplus[/tex]kerT U={u l T(u)=cu}


    2. Relevant equations



    3. The attempt at a solution
    Now before I start, just one quick question about ker T:
    U seems to be an eigenspace since T(u)=cu with c the eigenvalue.
    But that must mean that the kernel has 0 as the only element since
    0 is not an eigenvector so it can't be in the eigenspace, right?
    And, since T^2=cT, c cannot be 0 otherwise we would have T(0).
     
  2. jcsd
  3. Jun 17, 2009 #2

    tiny-tim

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    Hi evilpostingmong! :smile:

    (have a not-equal: ≠ and try using the X2 tag just above the Reply box :wink:)
    T2 = cT means T is a projection …

    so suppose eg V is R3, and T is the projection onto the x-y plane (with c = 1),

    then kerT is the z-axis. :wink:
     
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