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Kernel, Range, Basis (linear algebra)

  1. Dec 13, 2009 #1
    Hey all!
    I am working on this and got confused. Any help at all would be much appreciated!

    Determine the kernel and range of the transformation T and find a basis for each: T(x,y,z)=(x,y,z) from R3 to R2.

    I have found the kernel to be the set {(r, -r, 0)}.
    Range is R2.
    I"m not sure how to find the basis. Any ideas? I think the range is right but not sure.
    Thanks guys!!
     
  2. jcsd
  3. Dec 13, 2009 #2
    The transformation your have written maps R^3->R^3.
     
  4. Dec 13, 2009 #3
    Sorry the transformation is T(x,y,z) = ( x + y, z) Any ideas what the basis is?
     
  5. Dec 13, 2009 #4
    What do you mean basis of a transformation? A basis is something associated with a subspace.
     
  6. Dec 14, 2009 #5
    He means a basis for the kernel and a basis for the range.

    So you have found the range to be all of R^2, then you may give any basis for R^2, recall a basis is a set of vectors from the space such that they are linearly independent and span the space. There is a standard basis for R^2 you may use here...

    Similarly, you have characterized all the vectors in the kernel by r*(1,-1,0), where r is any number in R, think about the definition of spanning a space: a set of vectors spans a space if any vector from the space can be expressed as a linear combination of those vectors. So now can you tell me what is the basis you have found for the kernel?
     
  7. Dec 14, 2009 #6

    HallsofIvy

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    The kernel consists of all (x,y,z) such that T(x,y,z)= (x+y, z)= (0, 0). That is, x+y= 0 or y= -x and z= 0. Any vector in the kernel is of the form (x, -x, 0)= x(1, -1, 0) and so {(1, -1, 0)} is a basis for that one-dimensional vector space.

    The range consists of all numbers of the form (x+y, z). Since x and y can be any numbers, so can x+y. The range is all of R2 and so has {(1, 0), (0, 1)} as basis.

    Notice that the two dimensions, 1 and 2, add to 3, the dimension of R3.
     
    Last edited by a moderator: Dec 14, 2009
  8. Dec 14, 2009 #7
    Okay that makes sense. Thanks guys! The basis is the standard basis of {(1,0),(0,1)} And the range can be any number in R2.
    and yes the dim Ker(T) is 1, dim Range(T) is 2, dim Domain(T) is 3.
     
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