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(algebra) Proving subspaces- functions

by natalie:)
Tags: algebra help, help asap, spans, subspaces
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natalie:)
#1
Oct11-10, 06:21 PM
P: 4
1. The problem statement, all variables and given/known data

Is U={f E F([tex]\left|a,b\right|[/tex]) f(a)=f(b)} a subspace of F([tex]\left| a,b \right|[/tex]) where F([tex]\left| a,b \right|[/tex]) 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 [tex]\Theta[/tex]
- 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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Mark44
#2
Oct11-10, 06:38 PM
Mentor
P: 21,312
Quote Quote by natalie:) View Post
1. The problem statement, all variables and given/known data

Is U={f E F([tex]\left|a,b\right|[/tex]) f(a)=f(b)} a subspace of F([tex]\left| a,b \right|[/tex]) where F([tex]\left| a,b \right|[/tex]) 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 [tex]\Theta[/tex]
- 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 Quote by natalie:) View Post
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)?
natalie:)
#3
Oct11-10, 06:50 PM
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...

Mark44
#4
Oct11-10, 06:59 PM
Mentor
P: 21,312
(algebra) Proving subspaces- functions

Quote Quote by natalie:) View Post
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 Quote by natalie:) View Post

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 Quote by natalie:) View Post
I understand the point of whats happening but not the logic behind whats actually happening, if that makes any sense...
natalie:)
#5
Oct11-10, 07:12 PM
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!
Mark44
#6
Oct11-10, 07:30 PM
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P: 21,312
Quote Quote by natalie:) View Post
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 Quote by natalie:) View Post
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 Quote by natalie:) View Post
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 Quote by natalie:) View Post
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 Quote by natalie:) View Post

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 Quote by natalie:) View Post

Does the absolute value change anything in solving or does it just imply the functions x-values stay above the axis?
What absolute value?
natalie:)
#7
Oct11-10, 07:40 PM
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!
Mark44
#8
Oct11-10, 07:45 PM
Mentor
P: 21,312
Quote Quote by natalie:) View Post
Wow I was really not understanding but now really am!!
Cool!
Quote Quote by natalie:) View Post
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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