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Kernel/Range basis

  1. Apr 30, 2010 #1
    L: R^4 => R^3 is defined by L(x,y,z,w) = (x+y, z+w, x+z)

    A) Find a basis for ker L

    We can re write L(x,y,z,w) as x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0).
    I then reduced it to row echelon form

    We now have the equations X-W=0 , Y+W=0, Z+W=0.

    There are infinitely many solutions as X=W, Y= -W and Z=-W. So if we set W=1 we have

    the basis for the kernel=Vector(1,-1, -1,1)

    B) find a basis for range L

    L(x,y,z,w) = (x+y, z+w, x+z)

    We can re write L(x,y,z,w) as x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0).
    S= {(1,01) ,(1,0,0) ,(0,1,1),(0,1,0)} It spans L.

    To find the basis for L we set {x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0)} = 0,0,0

    I reduced it and the leading one's appear in the first 3 columns of the reduced form, the first 3 vectors in the original matrix became a basis for the range of L
    They are:
    Vector( {(1, (0, (1})
    Vector( 1, 0, 0})
    Vector( 0, 0, 1})

    C) verify theorem 10.7 (dim(KerL) + dim(rangeL) = dim V

    The dim can be viewed as the # of vectors in of the Ker/range.

    Given (dim(KerL) + dim(rangeL) = dim V we have 1+3=4, which is the number of dimensions in the original space (L(x,y,z,w)).
  2. jcsd
  3. Apr 30, 2010 #2


    Staff: Mentor

    What's your question?
  4. Apr 30, 2010 #3
    I forgot to type it. I believe A and B to be correct, but is my explanantion in C suffice?
  5. Apr 30, 2010 #4


    Staff: Mentor

    Sure, it's fine, and the other parts are fine also.
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