Hi everyone , can anyone of you help me with this question ?

My background in Algebra is weak

Thanks a lot guys

Let T : V -> W be a linear mapping and let {v1, v2, ..., vn} be a basis for V such that {vr+1,...,vn} is a basis for Null(T). Prove that {T(v1),...,T(vr) is a basis for the range of T.

clear explanation is appreciated.

First, let y be in the range of T. Then y= Tx for some x in V. Since {v1, v2, ..., vn} is a basis for V, x= a1v1+ a2v2+ ...+ anvn for some numbers a1, a2, ..., an. Then Tx= a1Tv1+ a2Tv2+ ...+ an Tvn. But T(vr+1)= 0, T(vr+2)= 0, ... T(vn)= 0 so y= Tx= a1T(v1)+ a2Tv2+ ...+ arTvr. That shows that T(vr) spans the range of T.

To show independence, assume that b1T(v1)+ b2T(v2)+ ...+ brT(vr)= 0.

Then T(b1v1+ b2v2+ ...+ brvr)= 0. Since {vr, vr+1, ..., vn} is a basis for Null(T), and b1v1+ b2v2+ ...+ brvr is in Null(T), we must have b1v1+ b2v2+ brvr= cr+1 vr+1+ ...+ cnvn. Show, using the fact that {v1, v2, ..., vn} is a basis, that all of b1, b2, ..., br, cr+1, ...cn must be 0.