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Linear Transformation questions about dimensions

  1. Feb 8, 2012 #1
    1. Say you have a linear transform from A to B, and where A has a higher dimension than B. How do you show that the kernel of the transform has more than one element (i.e. 0)? Also, if B has a higher dimension than A, then how to show that the transform isn't surjective?

    2. The attempt at a solution

    By showing that the kernel has more than the element 0, I want to show that the transform isn't injective. But I'm not quite sure how to get there just by using the fact that A has a higher dimension than B. Is that a good way(as in, not too complicated) of proving it? Any ideas?

    For the other part, it makes sense intuitively, since the basis of A will have less elements than the basis of B, so there shouldn't be a surjection. But how do you proceed from there to show that the image of A is a proper subset of B?
     
  2. jcsd
  3. Feb 8, 2012 #2
    Think of the theorem that says the dim(A)=dim(Ker(f)) + dim(Im(f)), where f:A->B is a linear morphism (transformation)
     
  4. Feb 8, 2012 #3
    Thanks! Now I feel really stupid for not considering Rank-Nullity before asking this....
     
  5. Feb 8, 2012 #4
    You're welcome. I feel stupid for not knowing that theorem has a name ...
     
  6. Feb 8, 2012 #5
    Do you know any bad math jokes related to this?
     
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