Linear transformations - Proving that a set generates the targe space

[arkanoid]

Homework Statement

Let $$A: E \rightarrow F$$ be a linear transformation between vector spaces (of any dimension) and let X be a subset of F with the following property (which is only a conditional):

IF $$X \subseteq Im(A)$$ THEN A is surjective. ... (*)

Prove that X is a generating set for F.

Homework Equations

E1)Im(A) = set of images of elements of E under A.
E2)A is surjective if and only if it transforms generating sets into generating sets.
E3)A has right inverse if it is surjective

The Attempt at a Solution

I assumed the antecedent of (*) since I think I can start the proof that way. Then I can assert that on one hand that A is surjective and using E2) I concluded that Im(A) is a generating set which is redundant.
I also tried to take all y in F that are in Im(A) but not in X (because I assumed that X is inside Im(A) ) and get the inverse image of these elements so I can map these into X. But then I realized that I don't have the freedom to do that (I can at most assign arbitrary values to members of a basis).

I tried to use the contrapositive of (*) and assume that there exists some w in F such that it is not the image of any v in E. then it follows that there exists some x in X such that it is not the image of any v in E. But this got me nowhere.

Please help with this problem. Thanks

 

[micromass]

Hmm, there's something fishy about this. With the conditional statement "if $$X\subseteq im(A)$$ then A is surjective". Which do they mean:

1) Take an A fixed and X fixed. If $$X\subseteq im(A)$$ for this specific X and A, then A is surjective.

or

2) Take X fixed. If $$X\subseteq im(A)$$ for any A, then this A is surjective.


I think they mean (2). Since (1) wouldn't make any sense. But they way you frazed the question, kind of suggest (1)...

 

[arkanoid]

hi micromass, what they mean is that X is given and fixed but the only thing they tell me about this set is that any linear transformation A whose image contains X (if there is any, remember it is a conditional) must fulfill that it is surjective. Eventually I have to prove that X having this property is a generating set of F.

Yes, it is (2) what they mean

 

[micromass]

OK, that's good. I'll also assume that you're working with $$\mathbb{R}$$ vector spaces, not that it matters a lot...


Now consider the vector space $$E=\mathbb{R}^X$$. Can you find a basis for this vector space? Can you find an easy linear transformation f:E-->F??

 

[arkanoid]

Excuse me, uhm is this E a particular case plugged in for the vector space E in the statement of the problem as to motivate a general answer?

For that E I'm afraid I cannot find a basis in general since X could be infinite and then $$\mathbb{R}^{\infty}$$ so I cannot find explicitly a basis (I've seen this in the book I read)

If X were finite then I could say that a basis would be the linear functionals that send one of the elements to 1 and the others to zero, represented as $$f_{x_i}$$. Then an obvious linear transformation from E to F would be the one that sends $$f_{x_i} \rightarrow x_i$$, let's call it A. Then the property that X has would lead to the desired result that X is a generating set because it is sending a generating set to X but since the image of A contains (actually is equal to) X then A is surjective therefore X is a generating set.

However, I have only proved it for this very specific case when E is of that form... also I assumed that X is finite, which is not granted.

I'm sorry I have to leave right now... I'll be back in the afternoon in around 8 hours, I will take your suggestion into account. Although right now I'm confused about it. I hope I can solve this problem that has been bothering me a lot (not really homework but I really feel annoyed that the book gives the following hint: "It is actually much easier than it looks")

thank you!

 

[micromass]

Excuse me, uhm is this E a particular case plugged in for the vector space E in the statement of the problem as to motivate a general answer?

Well, I assume that we can choose E arbitrary, since the A:E-->F could be arbitrary.

For that E I'm afraid I cannot find a basis in general since X could be infinite and then $$\mathbb{R}^{\infty}$$ so I cannot find explicitly a basis (I've seen this in the book I read)

Hmm. But for every vector space, we can always find a basis (so I'm not sure why the book says that). However, the book is correct in saying that we probably cannot explicitely find a basis. I mean, if the vector base is infinite dimensional, then there is a basis, but we cannot explicitely find it.

However, in the case $$\mathbb{R}^X$$, we can find an explicit basis...

If X were finite then I could say that a basis would be the linear functionals that send one of the elements to 1 and the others to zero, represented as $$f_{x_i}$$. Then an obvious linear transformation from E to F would be the one that sends $$f_{x_i} \rightarrow x_i$$, let's call it A. Then the property that X has would lead to the desired result that X is a generating set because it is sending a generating set to X but since the image of A contains (actually is equal to) X then A is surjective therefore X is a generating set.

Yes, this is correct. But the exact same proof is also true if X is infinite...

I'm sorry I have to leave right now... I'll be back in the afternoon in around 8 hours, I will take your suggestion into account. Although right now I'm confused about it. I hope I can solve this problem that has been bothering me a lot (not really homework but I really feel annoyed that the book gives the following hint: "It is actually much easier than it looks")

thank you!

Haha, "it's much easier than it looks". That's got to be the worst hint ever...

 

[arkanoid]

Hi! thanks for your reply...

I'm afraid I think the proof (as I described it above) doesn't go through for infinite X and moreover, I am asked to prove the statement for any vector spaces E and F, not just of the form $$E = \mathbb{R}^X$$.

the reason I think that the proof doesn't work for X infinite is that the set of functionals $$f_{x_i}, \hspace{5pt} i\in I$$ (with I being an infinite set that indexes X) is no longer a generating set but only linearly independent (I checked this in many books and the underlying reason for this is that infinite dimensional vector spaces are never isomorphic to their duals or double duals). So the argument of saying that a linear function A mapping this set into X to get that X is a generating set doesn't follow anymore.

I was actually thinking about that odd hint... maybe it's just that the solution is very tricky but simple :S

any additional help from anyone would be appreciated,
thanks :)