Just a quick question on Linear Transformations.

Let S = {v1, v2, ... , vn} be a set of vectors in the vector space V in R^n. The vectors in S are said to be linearly dependent if one of the vectors can be expressed as a linear combination of the others. If not then S is linearly independent.

Okay, I want S to be linearly independent. Done. Now, I want to tranform them under the linear transformation T:R^n -> R^m.

Now I have a new set of vectors R such that {T(v1), T(v2), T(v3), ... T(vn)}.

Is this set R linearly independent or dependent? How would you prove this?

You can try to solve the new set of vectors. If there's a solution other than a₁ = 0, a₂ = 0... for a₁*T(v1) + a₂*T(v2) ... = 0, then the set is linearly dependent. If the only solution to a₁*T(v1) + a₂*T(v2) ... = 0 is the trivial solution (all zeros), the set is linearly independent.
Someone else confirm? My linear algebra is a little rusty.

From the definition that if T:R^n -> R^m is a linear transformation (which it is in the above question) only if all vectors v, u and scalars c...
a) T(u + v) = T(u) + T(v)
b) T(cu) = cT(u)

From the question we have a linear transformation T and the set of vectors S = {v1, v2, v3, ..., vn} is a linearly independent set in R^n. The the set R = {T(v1), T(v2), ..., T(vn)} could not possibly be independent. I would say it is dependent, because relationships (a) and (b) hold. Can anyone give me an example of this or a proof? I am working on one.

Thanks.

Ah, I misunderstood your question. Hopefully I read it correctly this time.

The set R could be independent. Say you had v1 = (1,0), v2 = (0,1), and your transformation resulted in T(v1) = (2, 0, 0), T(v2) = (0, 2, 0). These vectors are still linearly independent.

So you are saying that the new set R is linearly independent? How sure?

Just say you had the set S = {v1, v2, v3} in R^3.
v1 = (1, 1, 1)
v2 = (1, 1, 0)
v3 = (1, 0, 0)
(all these vectors are linearly independent)
Let T: R^3 -> R^2 be a linear transformation such that
T(v1) = (1, 0)
T(v2) = (2, -1)
T(v3) = (4, 3)
We call R = {T(v1), T(v2), T(v3)}. Each vector in R is linearly independent and I claim that no matter what the transformation T is (so long as it is linear) and for any set of linearly independent vectors S, the resulting set R will always contain linearly independent vectors.

This last claim I think is true but I am probably wrong. Can anyone tell me I am right or wrong?

Cheers

I'm not saying R is independent or dependent, just that it's not necessarily dependent. Gotta run to a meeting.

Consider this: The function f(v)= 0, that takes EVERY vector into the 0 vector, is a linear transformation (prove that!).

If S is an independent set of vectors, then f(S) consists of the single vector {0}! What does that tell you?

How can you say that the vectors in R are independent? They are only 2-D and you have 3 of them.

I'll give you an example of coefficients that show the dependence:

a₁ = -10, a₂ = 3, a₃ = 1.

Any common multiple of them will of course also show dependence.

Let v be any vector in V. Since 0v = 0 we have f(0) = f(0v) = 0f(v) = 0. This preserves linear combinations hence f is linear.

Is that right?

___QUESTION___
If you have a linear transformation T : R^n -> R^m and you also have a set of linearly independent vectors S = {v1, v2, ..., vn} in R^n, then is the set of vectors R = {T(v1), T(v2), ..., T(vn)} linearly independent or dependent? I need this answered.
__

Okay bad choice of example combined with 3am sleepiness. You are totally right.

Good question. If S is an independent set of vectors then after the linear transformation is must contain the 0 vector? Is that what you are saying? Then the set of vectors in S that maps into 0 is the kernel of S. Does this help me work out the question above?

No, it's not at right! According to that proof, any function that takes the 0 vector into 0 is a linear transformation: and that's NOT true!

It is not enough to look only at f(0). For the function such that f(v)= 0 for all v, f(u+ v)= 0 and f(u)+f(v)= 0+ 0= 0 for f(u+v)= f(u)+ f(v) for ALL u and v. f(kv)= 0 and kf(v)= k*0= 0 so f(kv)= kf(v) for ALL vectors v and numbers k (not just k=0).

No, that's not what I'm saying. The point is that the set of "transformed" vectors consists ONLY of the 0 vector and so is NOT independent. If {v₁,v₂,...v_n} is a set linearly independent and f a linear transformation, the set {f(v₁),f(v₂),...f(v_n)} is NOT necessarily linearly independent. Whether the resulting set is independent or not depends upon the specific linear transformation.

By the way, you also said "Each vector in R is linearly independent". That doesn't make sense. Individual vectors are not "linearly independent". "Linear independence" depends on the SET.

As you can see I have difficulty understanding pure mathematics, it makes it harder when you don't have any real-life situations that you can apply it too (unlike applied maths).

I have re-grouped and come up with a summary of my thoughts on this matter. If you have some time could you read this and tell me if I have sorted everything out or not. Thanks HallsofIvy or anyone else.

Summary:
If S = {v1, v2,..., vn} (and is non-empty) is a set of vectors which can be expressed as:
c1v1, c2v2, ..., cnvn = 0
Then it has only one solution in which c1, c2,..., cn = 0.
If this is the only solution then S is a linearly independent set.
Such linearly independent sets are the basis vectors in R^3 ie.
e1 = (1,0,0)
e2 = (0,1,0)
e3 = (0,0,1)
Now suppose we have a linear transformation T from R^n to R^m. This can be defined as the function that takes all vectors in a set S (S being a linearly independent set (established above)) to another set R. ie.
T(v1, v2, ..., vn) => (u1, u2, ..., um)
Note the change from n to m in the subscripts.
Now since T is a linear transformation (or function) then the vectors in S = {v1, v2, ..., vn} AND the vectors in R = {u1, u2,...,un} are all linear ! If and only if (a) and (b) hold.
(a) T(u+v) = T(u) + T(v)
(b) T(kv) = kT(v)
This has to be correct!

If (a) and (b) do not hold then R is linearly independent.

Since we can express any set S = {v1, v2, ...,vn}

i)So if I can express S as a linear combination of vectors in which the only solution is 0. Then apply the linear transformation. The resulting set R has to be linear independent.

ii)If the solution is not 0 (there are other solutions) then the linear transformation may or may not result in a linear independent set R.

Oxymoron. I hope this helps:

We assume that the vectors v1, v2..., vn are linearly independent. This means that c1*v1 + c2*v2 + ... + cn*vn = 0 has only one solution: c1=c2=...=cn=0.

Okay. Let's now transpose these vectors by a linear transformation T, which gives us the following n (not m! You mix up the dimension of the space the set is in and the number of elements in the set) vectors: T(v1), T(v2),....T(vn).

The question now is: Are these vectors also linearly independent? Well look at the equation: d1*T(v1)+d2*T(v2)+....+dn*T(vn)=0. If this equation has only the zero solution d1=d2=....dn=0, then the transformed vectors are also linearly independent.

Due to the linearity of T, we can rewrite the equation:

d1*T(v1)+d2*T(v2)+....+dn*T(vn)=0 ===> T(d1*v1+....dn*vn)=0

From this equation we can conclude that the transformed vectors T(v1),T(v2),...,T(vn) are linearly independent if and only if T is invertible.

Proof:

===> Assume that T is invertible. Then the equation Tx=0 has only the solution x=0. Then from T(d1*v1+....dn*vn)=0 we have d1*v1+....dn*vn=0 and thus d1=d2=....=dn=0, due to linearly independency of the original vectors. Hence then the transformed vectors are linearly independent.
<=== Assume that the transformed vectors T(v1),T(v2),...,T(vn) are linearly independent. Then the equation T(d1*v1+....dn*vn)=0 has only one solution: d1=d2=...=dn=0. Since the vectors v1,v2...,vn span R^n , this means Tx=0 has only the solution x=0, which means that T is invertible.

Summarized: given a set of linearly independent vectors v1,v2,.....vn, and a linear transform T, the set of transformed vectors T(v1),T(v2),...,T(vn) is linearly independent if and only if T is invertible.

Well hope i got everything right

*edit*: the above only holds for m=n of course

Thankyou Pim for your answer.

In short, could I say the following:

To prove that if the transformation T:R^n -> R^m is linear and the set of vectors S = {v1, v2, ... , vn} is linearly independent set of vectors in R^n, then P = {T(v1), T(v2), ... , T(vn)} is a linearly independent set of vectors in R^m - - if and only if T is invertible.

If T is invertible then P contains linearly independent vectors due to the linearity of the transformation (what you wrote).

If T is not invertible then P must contain at least one vector which is linearly dependent on one of the other vectors in P.

I thought T was only invertible if n = m?? Which rules out the possiblility of P containing all linear independent vectors.

_Next question_

You have a linear transformation T. And you have P (as above) which IS a linearly independent set of vectors. However, this time I want P = Im(T) ie. P is a set of linearly independent vectors which are in the Image of T. Then is it true to say: {v1, v2, ..., vn} is also linearly independent?

Thanks

That last question may be a little confusing. Let me say it another way...

If the transformation T:R^n -> R^m is linear and {T(v1), T(v2), ... , T(vn)} is a linearly independent set of vectors in Im(T), then is {v1, v2, ... ,vn} a linearly independent set of vectors in R^n????

oxymoron: You are right, a transformation T can of course only be invertible if n=m. Not very bright of me Let me rephrase my conclusion:

Given a set of linearly independent vectors v1,v2,.....vn, and a linear transform T, the set of transformed vectors T(v1),T(v2),...,T(vn) is linearly independent if and only if the equation Tx=0 only has one solution: x=0. (in the case m=n this is equivalent with invertability). I also proved this in my previous post. Note that for m smaller than n, the equation Tx=0 always has more solutions. So only for m larger or equal to n we can have linear independency of the transformed vectors.

I will come back to your other question later. No time right now, but at first sight i would say that this also only holds if Tx=0 only has the solution x=0.

I'm afraid I can't even make sense out of this. First it makes no sense to say that "the vectors in S = {v1, v2, ..., vn} AND the vectors in R = {u1, u2,...,un} are all linear !" The adjective "linear" does not apply to vectors. If you meant "linearly independent" then, again, you can't talk about individual vectors being "linearly independent". It is a SET of vectors that is or is not linearly independent.
In any case, what everyone has been trying to tell you is: Given that set S is linearly independent and T is a linear transformation it does NOT follow that R is linearly independenent. That depends upon both T and the particular set S.

Finally, it makes no sense to say these things hold "if and only if (a) and (b) are true". (a) and (b) are the conditions that T be linear and you had already assumed that.

No, that is not precisely true. Let T be defined on R^3 by
T(x,y,z)= (x, y, 0). Then T is not invertible but if S= {(1,0,0), (0,1,0)} then the set R= {T(1,0,0),T(0,1,0)}= {(1,0,0),(0,1,0)} IS linearly independent. Of course, if S= {(1,0,0),(0,1,0),(0,0,1)} then R= {(1,0,0), (0,1,0), (0,0,0)} which is NOT linearly independent.

What is true is that, given independent set S, the set R will also be independent if and only if T RESTRICTED TO THE SPAN OF R is invertible.

Okay I was rushing, but of course, I mean S is a set of linearly independent set. I understand how frustrating it must be to read some of the garbage that I write. Even I don't understand why I write some of those things. But thanks anyway for pointing them out.

So I was leaving out the 'restricted to the span of R' bit. How would you prove that formally? Also, T is invertible if n = m. If n does not equal m AND T is restricted to the span of R. Is R still a set of linearly independent vectors?

Thankyou for clearing that up HallsofIvy.

I just need quick, formal proofs of the following two questions:

1. 2. Everytime I get close someone throws me a spanner. I just need a simple proof, nothing more than what it says in the question.

I really appreciate everyone who is trying to help me with this - it is great to be able to talk through things and learn from my mistakes. Keep it up!

Regards.

suppose that the v's are not linearly independent in R^n. then there is some linear dependence, say &Sigma;a_ivⁱ=0. act T on both sides of this equation, and you have a linear dependence relation among T(vⁱ), which contradicts the assumption that they are independent.

this one is not true. for example, if m is less than n, then there is no way that the T(vi) can be independent.