# Linearly Independence and Sets of Functions

1. Feb 6, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

3. The attempt at a solution
I don't think I'm really understanding this problem. Let me tell you what I know: A set is linearly independent if $a_1 A_1 +...+a_n A_n = \vec0$ for $a_1,...,a_n \in R$ forces $a_1 = ...=a_n = 0$. If $f,g,h$ take any of the $x_i \in S$, then one of the $f,g,h$ will map that element of S to zero, and no set containing $\vec0$ can be linearly independent.

Somehow, however, I feel like I'm supposed to be arriving at the opposite conclusion.

2. Feb 6, 2012

### HallsofIvy

Staff Emeritus
Okay, so these will be dependent if it is possible to have Af(x)+ Bg(x)+ Ch(x)= 0 for all x with at least one of A, B, C not equal to 0. Using the information given, $Af(x_1)+ Bg(x_1)+ Ch(x_1)= A(0)+ B(1)+ C(1)= B+ C= 0.$ Use the information given about $x_2$ and $x_3$ to get two more equations and solve for A, B, C.

It is certainly true that no set containing the 0 vector is independent but that is irrelevant to this problem since none of f, g, or h is the 0 vector.

Last edited by a moderator: Feb 6, 2012
3. Feb 6, 2012

### TranscendArcu

So, $\vec0 = Af(x_2) + Bg(x_2) + Ch(x_2) = A +C$ and $\vec0 = Af(x_3)+ Bg(x_3) + Ch(x_3) = A+ B$. So we have equations $A+B = B+C =A+C = 0$, which necessitates that $A=B=C=0$. Thus, the only way to get the set vector from a linear combination of f,g,h is with all-zero coefficients, meaning that the set is linearly independent.

4. Feb 6, 2012

### Bacle2

You may want to formalize things some more if you feel confused, by formally making the space of (Real, I assume)-valued functions, into a vector space, choosing a basis and then testing.

5. Feb 6, 2012

### TranscendArcu

Okay. I've made my discussion more "formalized" in my work. Thank you. Can I also get some help with this problem:

Here's my work so far:

I know that a linear transformation must have the property: Let T take elements of V to elements of W such that T(A) +T(B) + T(A+B) for all A,B in V. Thus, I said, let c,f be elements of S. I need (I'll use "ψ" as my special character indicating the transformation in the problem),

ψ(T)(c) + ψ(T)(f) = T(χc) + T(χf) = T(χc + χf) = ψ(T)(c+f), so it passes addition. (Is that right? I'm not sure.) I also need rT(A) = T(rA) for all r in R. I write,

ψ(T)r(c) = T(r(χc)) = rT(χc) = r ψ(T)(c).

I'm not really sure that I'm doing these correctly. It would be most helpful if you could tell me what in particular is wrong with the approaches I attempt. That way, I think I will know both why the right way is right, and why the wrong way is wrong. Thanks!

6. Feb 6, 2012

### Bacle2

Just to make sure: is L(Fun(S),R) the vector space of linear maps from Fun(S) into R?

If this is so, it may help re showing linearity.

And if you can show that your map is a linear bijection between vector spaces of the same

dimension, that shows you have an isomorphism.

Last edited: Feb 6, 2012
7. Feb 6, 2012

### Bacle2

Actually, ψ is described as being linear in L(Fun(S),ℝ), not in S.

But you cannot declare it to be linear; linearity follows from the properties
of L(Fun(S),ℝ ), as well as from the definition of ψ .

Last edited: Feb 7, 2012
8. Feb 7, 2012

### Deveno

this problem is more complicated then it looks. let me help you "un-wrap" it:

φ:L(Fun(S),R)→Fun(S)

so the input for φ is an element of L(Fun(S),R), such as T:

φ:T→φ(T).

now φ(T) is in Fun(S), which means it takes elements of S to elements of R:

φ(T):s→φ(T)(s) = T(χs),

where χs is the characteristic function of s

(this would make more sense to me if it were χ{s}, as i am used to seeing characteristic functions defined on sets).

to show φ is linear, you need to show that if T and T' are 2 elements of L(Fun(S),R):

φ(T+T') = φ(T) + φ(T')
φ(cT) = cφ(T).

these are equalities of functions, so to prove they are equal functions, you need to show that for every s in S:

φ(T+T')(s) = φ(T)(s) + φ(T')(s)
φ(cT)(s) = cφ(T)(s).

to evaluate the functions given above, you determine if:

(T+T')(χs)

and T(χs) + T'(χs) are equal,

and similarly for the second equation.