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Problem with this question(lenear algebra)

  1. Mar 2, 2009 #1
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 2, 2009 #2
    A) Your solution looks correct.

    B) You found correctly that Ker(T) consists of vectors of the form (x,0,0,0). This can can be written as Ker(T)=span{(1,0,0,0)}.
    Regarding the image of T: Your solution is correct, but one can also find a simpler basis consisting of standard basis vectors.

    C) Look at the system of linear equations you extracted from the matrix equation, there is a small error. Also, since T has a one-dimensional kernel there will be infinitely many solutions.
     
  4. Mar 2, 2009 #3
    regarding B i get 0*x=0
    on what basis i can state that its
    (x,0,0,0)
    ??

    where is the error in C
    ??
     
  5. Mar 2, 2009 #4
    Well, you are trying to find all possible values of x,y,z,t so that T((x,y,z,t))=0. You found that this means that 0=x*0, y=0, z=0, t=0. So x can be any number (0=x*0 is true for all x) and y,z,t have to be zero. So the solutions of the equation T((x,y,z,t))=0 are exactly the vectors of the form (x,0,0,0), where x is any number.

    The first equation, x+2y=2, should be y+2z=2.
     
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