Suppose w1,…,wn is a basis of W and V is finite-dimensional. Suppose T∈L(V,W). Prove that there exists a basis v1,…,vm of V such that all the entries in the first row of M(T) (with repsect to the bases v1,…,vm and w1,…,wn) are 0 except for possibly a 1 in the first row, first column.
(In this exercise, unlike 3c/3, you are given the basis of W instead of being able to choose a basis of W.)
If w1∈range T1 set v1 so Tv1=w1, otherwise (TODO)
then extend to a basis v1,…,vn of V. For each k>1 we have
T(vk)=ckw1+…
Meaning T(vk−ckv1) has no w1 component, so consider the list
(v1,v2−c2v1,…,vn−cnv1)
It's linearly independent because a1v1+⋯+an(vn−cnv1)=0 implies
(a1−a2c2−⋯−ancn)v1+a2v2+⋯+anvn=0
Since v1,…,vn are independent we have a2=⋯=an=0 meaning the original equation is a1v1=0 thus a1=0 as well. So our new list is a basis by 2.39, and we clearly have
T(a1v1+a2(v2−c2v1)+⋯+an(vn−cnv1))=a1w1+b2w2+⋯+bmwm
As desired, meaning M(T) with respect to the v1,…,(vn−cnv1) basis will have the first row be (1,0,…,0).
This proof is effectively a matrix algorithm style proof, but written in terms of linear maps.