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Not (null T) is a subspace of V?

  1. Jun 16, 2015 #1
    If T is a linear operator L(V,W) then can we say that all the vectors (in the vector space V) that T does not map to the zero vector (in the vector space W) form a subspace call it X?

    If a collection of vectors forms a subspace then they must satisfy closure under vector addition and scalar multiplication and there must also exist an additive identity. Taking this into consideration, it is clear that the elements of X are closed under addition (but multiplication not so sure), but its not clear (to me atleast) if W has an additive identity.

    Why is X closed?
    Suppose the set {x1,x2,...} is a basis for X then the action of T on any linear combination of this set will not lead to a zero vector in W (by linearity of T)

    However, if we can never get a zero vector in W by T on X then since T is linear we cannot have the scalar zero multiplying the elements of X since T(0*x)=0 (required for linearity) where x ε X. Hence the additive identity doesnt belong to X?

    If this doesnt work can someone suggest an alternative definition for X such that this would work?
    My present definition is X = { v ε V : Tv =\=0} .

    Thank you :) my motivation here is purely aesthetic i just wanted to write V=(null T)⊕ X

    Note: I wasnt able to write "is an element of" so i represented it by epsilon ε
  2. jcsd
  3. Jun 16, 2015 #2
    If i define X such that the intersection of X with null T is the zero vector then this definition works this leads to V = null T ⊕ X. But i am still interested in what went wrong with the earlier definition?
  4. Jun 16, 2015 #3


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    I suppose you simply need to declare that the zero vector is in all three spaces. Adding the zero vector into X should take care of your requirements for X to be a subspace.
  5. Jun 16, 2015 #4


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    The problem with your earlier definition was just that...it didn't include the zero vector. Every vector space includes the additive identity, and any linear operator acting on the zero vector must return zero, so you can be sure zero was not in X.
  6. Jun 16, 2015 #5
    Yeah, thank you :)
  7. Jun 16, 2015 #6


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    Often you see this referred to as the column space of the operator T.
  8. Aug 14, 2015 #7


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    In general, given subspace U of vector space, V, the set of all vectors that are NOT in U does NOT form a subspace.
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