# Linear transformation and indepence problem

• Jan 29th 2008, 10:40 PM
lllll
Linear transformation and indepence problem
I am having trouble with a problem and am in need of some help.
V & W are vector space, and T:V->W is linear.
T is 1 to 1 and S is a subset of V. Prove that S is linearly independent if and only if T(S) is linearly independent.

I have started the problem but am really uncertain of my approach. So far I have
S=(a1,a2,...,an)
V=(a1,a2,...,an,an+1,...,an+k)

Sum [biT(ai)] = 0 where i=1,2,...n
T[Sum bi(ai)] = 0

therefore there should exist some c such that

Sum [bi(ai)] = Sum ci(ai)

Is what I have done even correct?
• Jan 30th 2008, 06:48 AM
ThePerfectHacker
Quote:

Originally Posted by lllll
I am having trouble with a problem and am in need of some help.
V & W are vector space, and T:V->W is linear.
T is 1 to 1 and S is a subset of V. Prove that S is linearly independent if and only if T(S) is linearly independent.

Let $S = \{ a_1, ... a_n\}$ and suppose this set is linearly independent. Then $T(S) = \{T(a_1),...,T(a_n)\}$. Suppose there exists $c_1,...,c_n$ (in the field) such that $c_1T(a_1) + ... + c_nT(a_n) = 0$ thus $T(c_1a_1+...+c_na_n) = 0$. But since $T$ is one-to-one its kernel is the zero-vector thus $c_1a_1+...+c_na_n \implies c_1 = c_2 = ... = c_n$. Thus, $\{T(a_1),...,T(a_n) \}$ are linearly independent.
• Jan 30th 2008, 08:13 PM
lllll
Would it be a similar prove if S was a basis instead of a subset?