Question

Prove that T is one to one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W.

Answer 1

Answer with Step-by-step explanation:

Suppose T is one-one

Let S be a linearly independent subset of V

We want to show that T(S) is linearly independent.

Suppose T(S) is linearly dependent.

Then there exist $v_1,v_2,...v_n\in S$ and some not all zero scalars $a_1,a_2,....a_n$ such that

$a_1T(v_1)+a_2T(v_2)+a_3T(v_3)+...+a_nT(v_n)=0$

T is linear therefore,

$T(a_1v_1+a_2v_2+..+a_nv_n)=0$

T is one-one therefore

N(T)=0

$a_v_1+....+a_nv_n=0$

S is linearly independent therefore,

$a_1=a_2=...a_n=0$

It is contradiction.Hence, T(S) is linearly independent.

Conversely, Suppose that T carries linearly independent subset of V onto linearly independent subsets of W.

Assume that T(x)=0 if the set x is linearly independent

Then, by assumption we conclude that  {0} is  linearly independent but {0} is linearly dependent.

It is contradiction .Hence, the set {x} is linearly dependent which implies that x=0

It means N(T)={0}.Therefore, T is one- one