(algebra) Proving subspaces- functions

by natalie:)
Tags: algebra help, help asap, spans, subspaces
 P: 4 1. The problem statement, all variables and given/known data Is U={f E F($$\left|a,b\right|$$) f(a)=f(b)} a subspace of F($$\left| a,b \right|$$) where F($$\left| a,b \right|$$) is the vector space of real valued functions defined on the interval [a,b]? 2. Relevant equations I know in order for something to be a subspace there are three conditions: - existence of the zero vector $$\Theta$$ - closed under addition - closed under scalar multiplication 3. The attempt at a solution I tried to determine some boundaries for the function (not 1 to 1, all values will be positive, a and b can be any real number). I'm not really sure how to approach it ... i thought maybe if f(a)=0 then f(b) will also =0. But i don't really know how to prove that or the other two conditions. I would really appreciate someone explaining the steps of how to solve this. Thanks!!
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P: 21,312
 Quote by natalie:) 1. The problem statement, all variables and given/known data Is U={f E F($$\left|a,b\right|$$) f(a)=f(b)} a subspace of F($$\left| a,b \right|$$) where F($$\left| a,b \right|$$) is the vector space of real valued functions defined on the interval [a,b]? 2. Relevant equations I know in order for something to be a subspace there are three conditions: - existence of the zero vector $$\Theta$$ - closed under addition - closed under scalar multiplication 3. The attempt at a solution I tried to determine some boundaries for the function (not 1 to 1, all values will be positive, a and b can be any real number). I'm not really sure how to approach it ... i thought maybe if f(a)=0 then f(b) will also =0.
Don't put any restrictions such as these on your functions.
 Quote by natalie:) But i don't really know how to prove that or the other two conditions. I would really appreciate someone explaining the steps of how to solve this. Thanks!!
Clearly U contains the zero function.

Now, let f and g be any arbitrary functions in U, and c any scalar. What can you say about (f + g)(a) and (f + g)(b)?

What can you say about cf(a) and cf(b)?
 P: 4 Oh I understand we can just name two functions since U represents the set of functions... well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? i'm not really sure if i'm right about that. then c*f(a)=c*f(b).. I understand the point of whats happening but not the logic behind whats actually happening, if that makes any sense...
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P: 21,312
(algebra) Proving subspaces- functions

 Quote by natalie:) Oh I understand we can just name two functions since U represents the set of functions... well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? i'm not really sure if i'm right about that.
(f+g)(a) = f(a) + g(a) = f(b) + g(b) = (f + g)(b)
Can you give reasons for each of the = signs above?
What can you conclude about the function f + g?
 Quote by natalie:) then c*f(a)=c*f(b)..
Or better, (cf)(a) = c*f(a) = c*f(b) = (cf)(b)
Can you give reasons for each = above? What can you conclude about the function cf?
 Quote by natalie:) I understand the point of whats happening but not the logic behind whats actually happening, if that makes any sense...
 P: 4 Ok well... (f+g)(a)=f(a)+g(a) i have no idea i want to say distributive property? f(a)+g(a)= f(b)+g(b) because of the limitations in the question right? if they both have to come to the same solution.. f(a)=f(b) then g(a)=g(b). For some reason all i can think about is extreme value theorem which has nothing to do with this. The function f+g is... no idea. For scalar multiplication it makes sense that if you multiply any scalar by the function its the same as multiplying it inside the function c*f(a)=(cf)(a), and again i'm not really sure but think it has something to do with the f(a)=f(b)... Does the absolute value change anything in solving or does it just imply the functions x-values stay above the axis? Thank you for your patience!
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P: 21,312
 Quote by natalie:) Ok well... (f+g)(a)=f(a)+g(a) i have no idea i want to say distributive property?
No, this is the sum of two functions is defined. The distributive property shows how multiplication distributes over a sum of numbers. There is no multiplication in (f + g)(a).
 Quote by natalie:) f(a)+g(a)= f(b)+g(b) because of the limitations in the question right?
Sort of. We're assuming that f and g are in U, so f(a) = f(b) and g(a) = g(b). Adding g(a) to f(a) is the same as adding g(b) to f(b).
 Quote by natalie:) if they both have to come to the same solution.. f(a)=f(b) then g(a)=g(b).
No, one doesn't follow from the other. They both follow from our assumption that f and g are in U.
 Quote by natalie:) For some reason all i can think about is extreme value theorem which has nothing to do with this. The function f+g is... no idea.
What you should conclude is that when f and g are in U, then their sum, f + g, will also be in U. Another way to say this is that U is closed under addition.
 Quote by natalie:) For scalar multiplication it makes sense that if you multiply any scalar by the function its the same as multiplying it inside the function c*f(a)=(cf)(a), and again i'm not really sure but think it has something to do with the f(a)=f(b)...
There is nothing going on "inside the function." c*f(a) = (cf)(a) by the definition of scalar multiples of a function.

Given two functions f and g, you can create four new functions: f + g, f - g, f*g, and f/g. Besides these four using arithmetic operations, there is also the composition f o g, where (f o g)(x) = f(g(x)). And there's the other order, g o f.

 Quote by natalie:) Does the absolute value change anything in solving or does it just imply the functions x-values stay above the axis?
What absolute value?
 P: 4 Wow I was really not understanding but now really am!! Oh and i was confused at the notation i thought the square brackets were abs value bars but they're brackets Thank you so much for your help!
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P: 21,312
 Quote by natalie:) Wow I was really not understanding but now really am!!
Cool!
 Quote by natalie:) Oh and i was confused at the notation i thought the square brackets were abs value bars but they're brackets Thank you so much for your help!
You're welcome. I enjoy doing it.

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