Linear Algebra: Linear Transformation and Linear Independence

[b0it0i]

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 a lot
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?

 

[HallsofIvy]

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?

No, you have shown that 0v1+ 0v2+ ...+ 0vk= 0 but we knew that anyway!
Try an "indirect proof". If v1, v2, ..., vk are NOT independent then there exist
a1, a2,..., ak, NOT all 0, such that a1v1+ a2v2+ ...+ akvk= 0. What happens if you take T of both sides of that?

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

No, it doesn't. wi may not even be in vector space V.

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 a lot

There is nothing in here to suggest that either {w1, w2,..., wk} or {w1, w2,...,wk} is a basis only that they are independent. Try the way I suggested.

 

[b0it0i]

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?

 

[proton]

I'm a beginner to proofs myself, but your proof looks correct to me. I don't think you needed to use contradiction.

Here's my work, although I'm not 100% sure its correct either:

suppose f(c1v1 + c2v2 + ... ckvk) = c1w1 + c2w2 + ... ckwk = 0
Then, since (w1, ..., wk) is linearly independent, c1=c2=...ck=0

As T is linear, f(0) = 0 since W is a vector space?

Thus, c1v1 + c2v2 + ... ckvk = 0. From above, c1=c2=...ck=0. Thus v1, v2, ...vk is linearly independent

 

[HallsofIvy]

I'm a beginner to proofs myself, but your proof looks correct to me. I don't think you needed to use contradiction.

Here's my work, although I'm not 100% sure its correct either:

suppose f(c1v1 + c2v2 + ... ckvk) = c1w1 + c2w2 + ... ckwk = 0
Then, since (w1, ..., wk) is linearly independent, c1=c2=...ck=0

As T is linear, f(0) = 0 since W is a vector space?

Thus, c1v1 + c2v2 + ... ckvk = 0. From above, c1=c2=...ck=0. Thus v1, v2, ...vk is linearly independent

first, if you are given a linear tranformation, T, don't start talking about "f"!

Second, while it is certainly true that T(0)= 0, it is NOT always true that if T(v)= 0, then we must have v= 0! And that's the way you need.

 

[HallsofIvy]

I'm a beginner to proofs myself, but your proof looks correct to me. I don't think you needed to use contradiction.

Here's my work, although I'm not 100% sure its correct either:

suppose f(c1v1 + c2v2 + ... ckvk) = c1w1 + c2w2 + ... ckwk = 0
Then, since (w1, ..., wk) is linearly independent, c1=c2=...ck=0

As T is linear, f(0) = 0 since W is a vector space?

Thus, c1v1 + c2v2 + ... ckvk = 0. From above, c1=c2=...ck=0. Thus v1, v2, ...vk is linearly independent

first, if you are given a linear tranformation, T, don't start talking about "f"!

Second, while it is certainly true that T(0)= 0, it is NOT always true that if T(v)= 0, then we must have v= 0! The kernel of a linear transformation is not necessairily {0}. And that's the way you need.

 

[proton]

ok let me try it again:

suppose c1v1 + c2v2 + ... ckvk = 0. Taking T of both sides, we obtain
T(c1v1 + c2v2 + ... ckvk) = T(0) = 0 [since T(0) = 0 because T is linear?] Then
c1w1 + c2w2 + ... ckwk = 0
Then, since (w1, ..., wk) is linearly independent, c1=c2=...ck=0

Thus, for c1v1 + c2v2 + ... ckvk = 0, with c1=c2=...ck=0, S = {v1,v2,...vk} must be a linearly independent set

 

[matt grime]

[since T(0) = 0 because T is linear?]

If you don't understand whty T(0) is 0 then you should try to prove it.

 

[radou]

ok let me try it again:

suppose c1v1 + c2v2 + ... ckvk = 0. Taking T of both sides, we obtain
T(c1v1 + c2v2 + ... ckvk) = T(0) = 0 [since T(0) = 0 because T is linear?] Then
c1w1 + c2w2 + ... ckwk = 0
Then, since (w1, ..., wk) is linearly independent, c1=c2=...ck=0

Thus, for c1v1 + c2v2 + ... ckvk = 0, with c1=c2=...ck=0, S = {v1,v2,...vk} must be a linearly independent set

Looks correct.