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## Homework Statement

Let V and W be vector spaces, Let T: V --> W be linear, and let {w1, w2,..., wk} be linearly independent subset of R(T). Prove that if S = {v1,v2,...vk} is chosen so that T(vi) = wi, for i = 1, 2,...,k, then S is linearly independent

## Homework Equations

## The Attempt at a Solution

I have no idea where to start with this proof, I've been looking over past theorems to gather information, but I'm not sure how to connect them to show that

S is linearly independent. Can anyone give a hint or suggestion

Attempt 1:

Since T is linear, we know that

T(x + y) = T(x) + T(y)

T(ax) = aT(x)

Since

{w1, w2,..., wk} is linearly independent subset of R(T)

aw1 + bw2 + ... + cwk = 0

a = b = c = 0

T(v1 +...+ vk) = T(v1) +...+ T(vk) = w1 +...+ wk (and since linearly independent)

0w1 +...+ 0wk = 0,

so then i worked backwards

0 = 0w1 +...+ 0wk = 0T(v1) +...+ 0T(vk) = 0T(v1 +...+ vk) = T(0v1 +...+ 0vk)

and hence 0v1 + ...+ 0vk = 0

therefore S = {v1,v2,...vk} is linearly independent?

Attempt 2:

since R(T) is a subspace of W

and {w1, w2,..., wk} is a subset of R(T)

then span{w1, w2,..., wk} is a subset of R(T)

I was hoping to show that S is a basis, and hence S is linearly independent

but i couldn't get to that

T(vi) = wi

does that mean each vector, wi, can be written as a unique linear combination of vi

and hence vi is a basis? thus S is linearly independent?

i don't think my methods are correct

any suggestions would be helpful, thanks alot

NEW ATTEMPT:

Alrite, I think I got it

but when you said use an

"Indirect Proof", I used proof by contradiction and obtained this

Problem:

Let V and W be vector spaces, Let T: V --> W be linear, and let {w1, w2,..., wk} be linearly independent subset of R(T). Prove that if S = {v1,v2,...vk} is chosen so that T(vi) = wi, for i = 1, 2,...,k, then S is linearly independent

S = {v1,v2,...vk} is chosen so that T(vi) = wi, for i = 1, 2,...,k --> S is linearly independent

I assumed the negation

S = {v1,v2,...vk} is chosen so that T(vi) = wi, for i = 1, 2,...,k and S is Linearly Dependent

Since S is linearly dependent

a1v1 + a2v2 + ... akvk = 0

such that there exists a nonzero coefficient

then as you suggested, by taking T of both sides

T(a1v1 + a2v2 + ... akvk) = T(0)

and since T is linear

a1 T(v1) + ... + ak T(vk) = 0 [since T(0)=0 ]

then our other assumption, T(vi)=wi

implies that

a1w1 +...+ akwk = 0

Since there existed a nonzero coefficient,

that implies that

{w1, w2, ... , wk} is Linearly DEPENDENT

which contradicts the statement that {w1, w2, ... , wk} is Linearly Independent

thus the original statement is true

is this correct?

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