Linear Algebra-Subspace Functions

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Homework Statement



Consider w= {f [itex]\in[/itex] F([itex]\Re[/itex]|f(-x)=f(x) for all x [itex]\in[/itex]R
Use the subspace test to verify W is a subspace of F(R)

Homework Equations





The Attempt at a Solution



0 is in W obviously

for closure under addition:
(f+g)(x) = (f+g)(-x) = f(x) +g(x) = f(-x)+g(-x)

I am confused how to verify closure under scalar multiplication

af(x) = af(-x) = (af)(x)=(af)(-x)?

I am not sure how to do this please help thanks
 

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  • #2
Fredrik
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Homework Statement



Consider w= {f [itex]\in[/itex] F([itex]\Re[/itex]|f(-x)=f(x) for all x [itex]\in[/itex]R
Use the subspace test to verify W is a subspace of F(R)

Homework Equations





The Attempt at a Solution



0 is in W obviously

for closure under addition:
(f+g)(x) = (f+g)(-x) = f(x) +g(x) = f(-x)+g(-x)

I am confused how to verify closure under scalar multiplication

af(x) = af(-x) = (af)(x)=(af)(-x)?

I am not sure how to do this please help thanks
You should use the definitions of the addition and scalar multiplication operations. They are of course (f+g)(x)=f(x)+g(x) and (af)(x)=a(f(x)). So when you check closure under addition, it should look like this: (f+g)(x)=f(x)+g(x)=f(-x)+g(-x)=(f+g)(-x). Can you do the same for scalar multiplication?
 
  • #3
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I see that is more easy to understand. So it is (af)(x) = a(f(x))=a(f(-x))=(af)(-x)
 
  • #4
Fredrik
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Yes, that's exactly right.
 
  • #5
HallsofIvy
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You say "0 is in W obviously". Why is that obvious? If I were your instructor, I would want to be sure you understood exactly what "0" is in this case.
 
  • #7
Fredrik
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f(-0)=f(0)
This is very wrong. Looks like Halls was right to ask, and I was wrong not to.
 
  • #8
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then how do you do it? f(x)=0 f(-x)=0?
 
  • #9
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This is very wrong. Looks like Halls was right to ask, and I was wrong not to.

So f(x)=f(-x)=0?
 
  • #10
Fredrik
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You need to start by explaining which member of F(ℝ) is the zero vector. Once you have done that, you can start thinking about whether it's a member of the subset W or not.
 
  • #12
Fredrik
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No, you just have to correctly identify which one of the members of F(ℝ) is the zero vector. Once you have done that, it's easy to verify that it's a member of W.
 
  • #13
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I actually have no idea how to find which one is the member...I am lost now. And confused on what you mean by "members"
 
  • #14
Fredrik
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Are you perhaps more familiar with the term "element"? As in "2 is an element of the set {1,2,3}". "Member" is an alternative term for "element". So 2 is a member of the set {1,2,3}. The members (i.e. elements) of a vector space are also called "vectors".

Do you understand what a zero vector is? Suppose that X is a vector space, and suppose that z is a member of X. What is the property that z must have in order to be considered the zero vector of X.
 
  • #16
Fredrik
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X + z = x?
That's the right idea, but you need to include the words "for all". The correct statement is "For all x in X, we have x+z=x". So you do know what a zero vector is. Now, is there any member of F(ℝ) that has that property?

Did you type an uppercase X by accident, or did you mean something different from x+z=x?
 
  • #17
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I was typing on a phone so it autocorrects it to capitalized letter for the start of the sentence. f(x+0)=f(-x+0)?
 
  • #18
Fredrik
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f(x+0)=f(-x+0)?
No. You seem to be ignoring the definition of F(ℝ). How is F(ℝ) defined again?

Also, when you make a statement that's supposed to be part of a proof, you need to make sure that every variable is assigned a value, or is part of a "for all" or "there exists" statement. For example:

Bad: x+z=x

Good: Let X be a vector space. Let z be the zero vector of X. For all x in X, we have x+z=x.
 
  • #19
micromass
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Maybe you should start by actually giving some examples of elements in the "vector space"?? Right now, the definition is pretty abstract. So can you give two or three examples of elements?
 
  • #20
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I actually hav eno clue what the definition is, I kind of skipped the entire lecture on the vector space part, all I know it is similar with subspaces and etc. So I guess I have to just let this question go with marks taken off because it is due in 5 hours and its 4 am over here. Linear Algebra is more abstract than I like.

I need sleep. THanks for the help. I will come back tomorrow to try to understand the problem. I will just put random guesses for the question for now :P.
 
  • #21
Fredrik
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Hopefully this experience will at least make you realize that it's futile to try to prove a statement without using the definition of the terms and notations in the statement. You were supposed to prove a statement about a specific vector space denoted by F(ℝ). (You wrote ##F(\Re)## in post #1, but ℝ is the standard notation for the set of real numbers). So you absolutely have to use the definition of F(ℝ). If you don't, then there's no reason to think that whatever you have managed to prove has anything to do with F(ℝ).

By the way, I have answered two very similar questions in the last week, and both of those guys made the exact same mistakes you did: 1. They ignored the definition of the vector space they were working with. 2. They made statements about variables without assigning them values or saying "for all" or "there exists". 3. When they were supposed to verify that the zero vector was in the subset, they started considering stuff like f(0).

1 and 2 are probably the two most common mistakes made by people who are just starting out with proofs. 1 is a huge mistake, pretty much the biggest one you can make next to assuming that the statement you want to prove is true. 3 is a mistake that you wouldn't make if you hadn't already made mistake 1.
 
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  • #22
micromass
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Being able to solve problems like this is pretty important. So you will probably not get good marks on this question, but I still encourage you to solve this question. You should absolutely be able to get this. Maybe go to your office hours and tell the professor that this kind of question is troubling you?
 
  • #23
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Well at the point where the function touches the y axes so x=0, then no matter what the sign is both f(x)=f(-x)= the same thing. So does it mean that the vertex point at the origin satisfies this?
 
  • #24
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Well at the point where the function touches the y axes so x=0, then no matter what the sign is both f(x)=f(-x)= the same thing. So does it mean that the vertex point at the origin satisfies this?

Forget the f(x) = f(-x) for a moment. What function f is such that, if g is a function, g(x) + f(x) = g(x)?

You're making this a lot more complicated than it really is.
 
  • #25
Fredrik
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What function f is such that, if g is a function, g(x) + f(x) = g(x)?
This equality should be g+f=g. f and g are functions. f(x) and g(x) are real numbers.
 

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