Linear Algebra-Subspace Functions

[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)?

 

[micromass]

Indeed, so we have the function $f(x)=0$ for all $x$. Does this function satisfy $f(x)=f(-x)$?

 

[FinalStand]

yes. f(x)+0(x) = f(-x) + 0(x)?

 

[Mark44]

yes. f(x)+0(x) = f(-x) + 0(x)?

You don't seem very certain. micromass's question was this:

If f(x) = 0 for all real x, is f(-x) = f(x) for all real x? There is no addition involved.

 

[Fredrik]

yes. f(x)+0(x) = f(-x) + 0(x)?

When f is defined by f(x)=0 for all x in ℝ, as in micromass's post, then yes, we have f(x)=f(-x) for all x. So the equality in the quote above holds. But I'm not sure how the conversation drifted into this.

You seem to understand that there's an f in F(ℝ) such that f+g=g+f=g for all g in F(ℝ). You also seem to understand that an f in F(ℝ) such that f+g=g+f=g for all g in F(ℝ) is the zero vector of F(ℝ). You seem to understand that the f in F(ℝ) defined by f(x)=0 for all x in ℝ has this property. But you still don't seem to have realized that this specific f is the zero vector.

In other words, the f defined by f(x)=0 for all x in ℝ, is the zero vector of F(ℝ), and can therefore be denoted by 0.

Now all you have to do to find out if this specific f is in W, is to determine if f(x)=f(-x) for all x in ℝ.

 

[FinalStand]

I thought f(x) was the vector ? So do I have to say f(x) = 0 is the zero vector and belongs to the subspace since f(x)=f(-x)? I am not exactly sure how to answer this question. How is f a vector? I am so confused, this is so different from what we learned before I took this course. It is more abstract than I like it to be.

 

[Fredrik]

...what is the point on giong over the definitions? And what's the point of proofing this? I think it is all rubbish.

Linear algebra is along with the basics of calculus the most important topic in mathematics. It's used extensively in all sorts of applications of mathematics from quantum mechanics to computer graphics. It's also used in other areas of mathematics. It's a foundation on which many other things are built.

I would define Linear algebra as the subset of the mathematics of linear maps between vector spaces that can be dealt with without using point-set topology (limits of sequences, continuity of functions, etc.) (An equally reasonable but slightly different definition is that it's the mathematics of linear maps between finite-dimensional vector spaces). If you don't know how to check if a given set with an addition operation and a scalar multiplication operation is a vector space, then you will find everything else in linear algebra and its applications very hard or impossible to understand.

So this is as far from being pointless/rubbish/useless/worthless as anything in mathematics can get.

 

[FinalStand]

i was frustrated sorry about that

 

[FinalStand]

So my last response what do you have to say in the answer? DO you just have to say the zero vector is in f(x)=f(-x) as f(x)=0? I just don't know how to shwo your work? As I am not good with showing every step, like you said you have to say "for all x" or w.e. I lost all my marks on communications in Calculus as well because I am nto good at stating variables and working with them.

 

[Fredrik]

I thought f(x) was the vector ? So do I have to say f(x) = 0 is the zero vector and belongs to the subspace since f(x)=f(-x)? I am not exactly sure how to answer this question. How is f a vector? I am so confused, this is so different from what we learned before I took this course. It is more abstract than I like it to be.

f(x) is not a vector, it's a number in the range of the function f, which is a vector. f is called a vector because it's a member of a vector space.

You really need to include the words "for all" where it's appropriate. Note e.g. that when I wrote f(x)=f(-x) at the end of post #54, I made it clear that I was talking about a specific f (the zero vector) and all x. If I had only written "f(x)=f(-x)", that wouldn't have been clear.

The exact statement "f(x)=0 is the zero vector" doesn't really make sense, but I know what you mean. This is a good way to say it: "The function $f:\mathbb R\to\mathbb R$ defined by f(x)=0 for all x in ℝ, is the zero vector of F(ℝ)". It's also OK to just say that the zero vector of F(ℝ) is the function that takes every real number to 0.

To prove that this function is the zero vector, you must pick a notation for it, for example f, and then show that this f satisfies f+g=g for all g in F(ℝ). This is how you do that: Let g be an arbitrary member of F(ℝ). Since f and g both have the domain ℝ, so does f+g. For all x in ℝ, we have (f+g)(x)=f(x)+g(x)=0+g(x)=g(x). This implies that f+g=g.

To prove that this f is in W, all you have to do is to prove that for all x in ℝ, we have f(x)=f(-x). This is how you do that: Let f be the zero vector of F(ℝ). For all x in ℝ, we have f(x)=0=f(-x).

Since it's standard to denote the zero vector of any vector space by 0, I think it's also fine to simplify that last proof to this: For all x in ℝ, we have 0(x)=0=0(-x). This does however look a bit weird, since the first and the last 0 denote the zero vector while the one in the middle denotes the number 0.

Yes, this is pretty abstract. But things will get a lot more abstract as you study more math. This is only the beginning. You just have to get used to it.

 

[FinalStand]

Ok, I was confused on what you guys meant by "vector" I thought the entire f(x) is considered a vector. and when it comes to the zero vector I was confused on whether the "f(x)" is the vector or the (x), so I was completely off. If this continues I think I am going to kill my brain, maybe I need to go take the lectures to make these easier.