Suppose V is finite-dimensional and T:V→W is a surjective linear map of V onto W. Prove that there is a subspace U of V such that T∣U is an isomorphism of U onto W. (Here T∣U means the function T restricted to U.)
Proof 1
From 2.34 we can find a subspace U of V such that U⊕null T=V.
Clearly null T∣U={0} since being a direct sum means U∩null T={0}.
And obviously range T∣U=range T=W since the nullspace doesn't contribute anything.
Making the last part rigorous, each v∈V can be uniquely written v=vu+vn for vu∈U, vn∈null T meaning if w=Tv then w=Tvu so the ranges are the same.
Proof 2
Let v1,…,vn be a basis of V and let wk=Tvk. we know span(w1,…,wn)=W so we can remove vectors (by 2.31) to get a a basis wk1,…,wkm of W then let U=span(vk1,…,vkm).
Obviously T∣U is surjective since Tvkj=wkj and wk1,…,wkm is spanning. To see injectivity use the independence of wk1,…,wkm
T(a1vk1+⋯+amvkm)=a1wk1+⋯+amwkm=0⟹a1=⋯=am=0
todo: come up with a proof that uses results from 3d