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?(adsbygoogle = window.adsbygoogle || []).push({});

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 {x_{1},x_{2},...} 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

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# Not (null T) is a subspace of V?

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