1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Kernel/Range basis
  1. Kernel, Basis, Rank (Replies: 10)