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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 = 0T(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 thatT(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 thisProblem:
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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