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Homework Help: Linear transformation and Ker(T)

  1. Nov 15, 2005 #1
    Hi, suppose I have a linear transformation T and Ker(T) consists of only the zero vector. Then is it true that a basis for Ker(T) consists of no vectors and is of dimension zero? I would like these technicalities to be clarified. Any help would be good thanks.
  2. jcsd
  3. Nov 15, 2005 #2


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    Yes, the "subspace" consisting only of the 0 vector has dimension 0.
  4. Nov 15, 2005 #3
    Ok thanks HallsofIvy.

    Just one more thing, if a transformation is injective then it is not necessarily onto? I ask this because I would like to know if in determining whether a transformation is invertible, if it is sufficient to look at whether or not it is injective.
  5. Nov 15, 2005 #4
    Correct. It's very easy to find an example of an injection which is not surjective.

    Sometimes it is (such as when the transformation is between two spaces with the same (finite) dimension).
  6. Nov 15, 2005 #5
    Thanks for quick reply Muzza.
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