Linear Algebra-Subspace Functions

[FinalStand]

Homework Statement

Consider w= {f \in F(\Re|f(-x)=f(x) for all x \inR
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

 

[Fredrik]

Homework Statement

Consider w= {f \in F(\Re|f(-x)=f(x) for all x \inR
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?

 

[FinalStand]

I see that is more easy to understand. So it is (af)(x) = a(f(x))=a(f(-x))=(af)(-x)

 

[Fredrik]

Yes, that's exactly right.

 

[HallsofIvy]

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.

 

[FinalStand]

f(-0)=f(0)

 

[Fredrik]

f(-0)=f(0)

This is very wrong. Looks like Halls was right to ask, and I was wrong not to.

 

[FinalStand]

then how do you do it? f(x)=0 f(-x)=0?

 

[FinalStand]

This is very wrong. Looks like Halls was right to ask, and I was wrong not to.

So f(x)=f(-x)=0?

 

[Fredrik]

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.

 

[FinalStand]

So I have to show examples?

 

[Fredrik]

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.

 

[FinalStand]

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"

 

[Fredrik]

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.

 

[FinalStand]

X + z = x?

 

[Fredrik]

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?

 

[FinalStand]

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)?

 

[Fredrik]

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.

 

[micromass]

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?

 

[FinalStand]

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.

 

[Fredrik]

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.

 

[micromass]

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?

 

[FinalStand]

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?

 

[Mark44]

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.

 

[Fredrik]

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.

 

[Fredrik]

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?

The 0 you're talking about is just a member of the domain of f. (And what is f anyway?) This has nothing to do with the zero vector of F(ℝ). You really need to look at the definition of F(ℝ).

I also second micromass' recommendation that you should try to think of a few examples of members of F(ℝ), just to make sure that you know what sort of things are considered "vectors" in this problem.

 

[micromass]

This equality should be g+f=g. f and g are functions. f(x) and g(x) are real numbers.

Or he should have said $g(x) + f(x) = g(x)$ for all x.

 

[micromass]

FinalStand, I think it's easier if you first think about this:

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?

So instead of trying to find the "0" element or trying to prove anything, I would first focus on finding a few examples of elements in your space.

 

[Mark44]

What function f is such that, if g is a function, g(x) + f(x) = g(x)?

This equality should be g+f=g.

It makes sense either way. For you to insist that the equality should be as you wrote, is a bit pedantic, IMO.

f and g are functions. f(x) and g(x) are real numbers.

As micromass points out, the equation should hold for all real x. That's what I meant, but my thought that was that the OP would understand what I meant.

 

[FinalStand]

Where is a good site I can read the definition? Because I skipped all classes on vector spaces and my prof doesn't post notes online. And the prof only took off 0.5 points off for this question.

And I really have no clue how this works. Now I don't even know what the point of (af)(x)=a(f(x)) is

 

[FinalStand]

F(x)+-f(-x)=0

 

[Mark44]

Where is a good site I can read the definition?

Don't you have a textbook? Any linear algebra textbook would have a definition of "vector space."

Because I skipped all classes on vector spaces and my prof doesn't post notes online. And the prof only took off 0.5 points off for this question.

And I really have no clue how this works. Now I don't even know what the point of (af)(x)=a(f(x)) is

 

[Fredrik]

Where is a good site I can read the definition? Because I skipped all classes on vector spaces and my prof doesn't post notes online. And the prof only took off 0.5 points off for this question.

And I really have no clue how this works. Now I don't even know what the point of (af)(x)=a(f(x)) is

The book you're using for this course must include a definition of the term "vector space". You can also look up this definition on Wikipedia. For this problem, you also need the definition of the vector space F(ℝ). The definition is almost certainly this:

Let F(ℝ) be the set of all functions from ℝ into ℝ. Now this just defines the set. To turn this into a vector space, we must define an addition operation and a scalar multiplication operation. The operations that your book/teacher had in mind are almost certainly defined as follows. Addition is defined like this: For all f,g in F(ℝ), we define f+g by
$$(f+g)(x)=f(x)+g(x)$$ for all x in ℝ. Scalar multiplication is defined like this: For all f in F(ℝ) and all a in ℝ, we define af by
$$(af)(x)=a(f(x))$$ for all x in ℝ.

Now you can easily verify that F(ℝ) is a vector space. (Strictly speaking, it's the triple (F(ℝ),addition,scalar multiplication) that should be called a vector space, but it's standard to abuse the terminology by referring to the set F(ℝ) as a vector space). A key step in this verification is to note that there's a z in F(ℝ) such that x+z=z+x=x for all x in F(ℝ). This unique z is denoted by 0 and is called the zero vector.

 

[FinalStand]

I know that but for f(x)=f(-x) can't we just add the x of the opposite sign then it givesyou 0? And no I did not buy the textbook because I never needed textbooks for math..well until maybe now.

 

[Mark44]

I know that but for f(x)=f(-x) can't we just add the x of the opposite sign then it givesyou 0?

No. All this equation tells you is that the graph of f is its own reflection across the y-axis. Many functions satisfy this relationship. Earlier in the thread micromass asked you if you could come up with two or three examples of functions that satisfy the equation above.

And no I did not buy the textbook because I never needed textbooks for math..well until maybe now.

I don't know what to tell you. If you can't be bothered to go to the class, or make the effort to buy the textbook, I don't know how much good we can do for you.

 

[FinalStand]

I give up. Thanks for the help

 

[Fredrik]

I know that

What does "that" refer to? Do you mean that you already knew that F(ℝ) is defined that way? If so, why have you still not realized that the zero vector is a function?

I suspect that there's also something lacking in your understanding of the concept of "function". Do you know e.g. what it takes for two functions to be equal? What do they have to have in common to be equal?

but for f(x)=f(-x) can't we just add the x of the opposite sign then it givesyou 0?

I don't even know what this means. What is f? Are you saying that this holds for all f in F(ℝ)? Are you saying that there exists an f in F(ℝ) such that this holds? Is f some specific member of F(ℝ), and you're saying that this holds for that f? What do you mean by "add the x of the opposite sign"? What do you mean by "gives you 0"? What 0? The number or the vector?

And no I did not buy the textbook because I never needed textbooks for math..well until maybe now.

You obviously need to rethink that strategy. It can't possibly work for very long.

 

[FinalStand]

Could it be (f+(-f))(x)=0 then f(-x)+ (-f(-x))=0? Can you guys tell me what the answer is so maybe I can work it out?

 

[Mark44]

Could it be (f+(-f))(x)=0 then f(-x)+ (-f(-x))=0? Can you guys tell me what the answer is so maybe I can work it out?

You seem to be fixated on this relatively minor point. What you appear to be doing is making random manipulations of this formula that describes which functions are in the set, without understanding what things are actually in the set. For example, is f(x) = x² + x + 1/2 in W?

Any function in the set must satisfy f(x) = f(-x), but that is only a characteristic of the members of the set, and is not the definition of any of them.

 

[FinalStand]

no that is not in the set...for example 1 would not work. You guys kept on repeating the same thing and I do not understand that thing. I don't know any thing and I have no idea what thing you are talking about either. I think I need to go drop this course or drop myself down a building and suicide.

 

[FinalStand]

And it does not touch the x-axis therefore it will not work. But I still don't see the point of this. I think the examples and the definitions are irrelevant. So what if it says x+y=x then why do we still need to proof that it is in f(x)=f(-x) ? of course there is an x that f(x+y=x) = f(-x+y=-x) what is the point on giong over the definitions? And what's the point of proofing this? I think it is all rubbish. i gave up on math I will change my major then. Screw myself with a million screw drivers.

 

[Mark44]

no that is not in the set

Are you referring to my example of f(x) = x² + x + 1/2? Why isn't it in the set? Show me your thinking.

...for example 1 would not work.

Actually, g(x) = 1 IS in set W.

You guys kept on repeating the same thing and I do not understand that thing. I don't know any thing and I have no idea what thing you are talking about either. I think I need to go drop this course or drop myself down a building and suicide.

And it does not touch the x-axis therefore it will not work.

What does this have to do with anything? There is no requirement that an arbitrary member of set W has to touch the x-axis.

But I still don't see the point of this. I think the examples and the definitions are irrelevant. So what if it says x+y=x then why do we still need to proof that it is in f(x)=f(-x) ? of course there is an x that f(x+y=x) = f(-x+y=-x) what is the point on giong over the definitions? And what's the point of proofing this? I think it is all rubbish. i gave up on math I will change my major then. Screw myself with a million screw drivers.

 

[Mark44]

Dropping the class is probably the best course of action for you, based on your obvious lack of interest in the course.

 

[FinalStand]

Believe it or not I do have a good mark in it. But what is the zero vector? I know that x+y=x. But for f(x)=f(-x) what is a zero vector for this? f(x)+g(x)=f(-x)? so g(x)=0? or does the value x in g(x) is 0? How is 1 in f(x)=f(-x) when your example is : x^2 + x + 1/2? f(1)=2.5 and f(-1)=0.5? clearly f(x)!=f(-x)? Do we have to prove this by induction or what?

In this thing which one is the vector the funciton itself or the value x? x^2+1 is not in the space I know that since f(x)=f(-x) for all x, but 0+1=1 which is not 0. Now, I love math, but these proofs are useless and worthless. I rather die than drop this course actually. But I probably end up being depressed and jump off a building.

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[FinalStand]

Are you referring to my example of f(x) = x² + x + 1/2? Why isn't it in the set? Show me your thinking.
Actually, g(x) = 1 IS in set W.


What does this have to do with anything? There is no requirement that an arbitrary member of set W has to touch the x-axis.

Where the hell did you get g(x)!? are you freaken blind? or mentally retarded like me? This is stupid.

 

[FinalStand]

f(-x) = 0 = -f(x) = f(-x)??

 

[micromass]

If I define the function $f(x) = x^2 +1$, then does this satisfy $f(x) = f(-x)$ for all $x$??

If I define the function $f(x) = x^2 + x +1$, then does this satisfy $f(x)=f(-x)$ for all $x$?

 

[FinalStand]

WHy does that matter!? Why are we looking at examples? no the second one does not satisfy all x. it is not a vectorspace at all. :). But how does this matter? I want ot know how to proof that whether 0 is in the space or not.

So (f+0)(x)=f(x)+0(x)=f(x)=f(-x) work?

 

[micromass]

WHy does that matter!? Why are we looking at examples? no the second one does not satisfy all x. it is not a vectorspace at all. :). But how does this matter? I want ot know how to proof that whether 0 is in the space or not.

So (f+0)(x)=f(x)+0(x)=f(x)=f(-x) work?

No, it doesn't work.

I'm doing examples to get you familiar with the space you're working in.

We are looking for a function $f$ such that $f+g=g$ for all functions $g$. So $f(x) + g(x) = g(x)$ for all $g$ and $x$. What function $f$ satisfies this?

 

[FinalStand]

0(x)?