Suppose that V is a finite-dimensional vector space and that T∈L(V,W). Prove that there exists a subspace U of V such that U∩null T={0} and range T={Tu:u∈U}.
Let n1,…,nk be a basis for null T then extend this to a basis n1…,nk,u1,…,ur of V using 2.33. Let
U=span(u1,…,ur)
Clearly U∩null T={0}, and range T={Tu:u∈U} since the nullspace components disappear by linearity
T(v)=T(a1n1+⋯+aknk+b1u1+⋯+brur)=T(b1u1+⋯+brur)=T(u)
I was tempted to try the set U′={v∈V:Tv=0}∪{0} but this doesn't work as it isn't closed under addition. Specifically notice that n1+u1∈U and n1−u1∈U but their sum n1 isn't. Having Tv=0 isn't enough, we need v to contain no nullspace component.