Why Does Adding Two Polynomials from a Set Not Always Result in a Vector Space?

[kq6up]

I am going through section 14 in chapter 3 in Mary Boas' "Mathematical Methods for the Physical Sciences 3rd ed"

Example 1 is clear, but this line is confusing in example 3 (I don't understand why it shows that these are not an element of the set). I have pasted example 1 for reference below.



"Modify Example 1 to consider the set of polynomials of degree ≤ 3 with f(1) = 1. Is this a vector space? Suppose we add two of the polynomials; then the value of the sum at x = 1 is 2, so it is not an element of the set. Thus requirement 1 is not satisfied so this is not a vector space. Note that a subset of the vectors of a vector space is not necessarily a subspace."

For reference, here is Example 1:

Example 1. Consider the set of polynomials of the third degree or less, namely functions of the form f(x) = a0 +a1x+a2x2 +a3x3. Is this a vector space? If so, find a basis. What is the dimension of the space?
We go over the requirements listed above:
1. The sum of two polynomials of degree ≤ 3 is a polynomial of degree ≤ 3 and so is a member of the set.
2. Addition of algebraic expressions is commutative and associative.
3. The “zero vector” is the polynomial with all coefficients ai equal to 0, and adding it to any other polynomial just gives that other polynomial. The additive inverse of a function f(x) is just −f(x), and −f(x) + f(x) = 0 as required for a vector space.
4. All the listed familiar rules are just what we do every time we work with algebraic expressions.
So we have a vector space! Now let’s try to find a basis for it. Consider the set of functions: {1, x, x2, x3}. They span the space since any polynomial of degree ≤ 3 is a linear combination of them. You can easily show (Problem 1) by computing the Wronskian [equation (8.5)] that they are linearly independent. Therefore they are a basis, and since there are 4 basis vectors, the dimension of the space is 4.

Thanks,
Chris Maness

 

[Matterwave]

The set specifically requires that f(1)=1. Elements f(x) with f(1)=2 are not in the set by definition. Adding two functions who has f(1)=1 will yield a function with f(1)=2 and so the sum is not in the set. Is that clear?

 

[kq6up]

The set specifically requires that f(1)=1. Elements f(x) with f(1)=2 are not in the set by definition. Adding two functions who has f(1)=1 will yield a function with f(1)=2 and so the sum is not in the set. Is that clear?

So you are saying it is not a part of the set because it is made up of the set. Correct?

Chris

 

[Matterwave]

So you are saying it is not a part of the set because it is made up of the set. Correct?

Chris

What? It is not part of the set because the set is defined not to include this element.

Maybe a simpler example is better. Say you have a set {1,2,3,4}. Are all the sums of members of this set included in this set? What about 4+1? Is 4+1 in this set?

 

[kq6up]

I meant it can't be part of the set because it is made up from the elements of the set, and I see from your clear example that this is the case.

 

[Matterwave]

I meant it can't be part of the set because it is made up from the elements of the set, and I see from your clear example that this is the case.

I don't see why this would be so... in the set {1,2,3,4}, 1+2=3 IS part of the set. 4+2 ISN'T part of the set. But whether an element is in a set or not depends on the set, not on whether that element is "made up from elements of the set" whatever you mean by that.

 

[kq6up]

Ok, I see it is simply by definition. Since it is not defined in the set, it is not part of the set. It seems obvious now.

I also understand the point you made:

The set specifically requires that f(1)=1. Elements f(x) with f(1)=2 are not in the set by definition. Adding two functions who has f(1)=1 will yield a function with f(1)=2 and so the sum is not in the set. Is that clear?

I am really having difficulty with this section in the book. I understood the group theory section right before it, and the whole rest of the LA chapter. I get to this section (which is the last section that is not a review), and it is like hitting a brick wall. What could I be lacking that causes this group/set stuff to make me feel obtuse? I don't normally struggle with math concepts to the point where I feel like I need tutoring. I have actually never used a math tutor, but I feel I might need to hire one for this section.

Thanks,
Chris Maness

 

[ModestyKing]

It's just a definition of a vector space and subspaces.

In a vector space, you must be able to add two elements of the set, given some restrictions/requirements, and still end up with a member of the set. If you have two polynomials f(x) and g(x), where f(1) = 1 for both of them, and you add them, then that polynomial at 1, p(1) = (f(1) + g(1)) = 1 + 1 = 2. It's as easy as 1 + 1 ;)

So that requirement that f(1) = 1 turns it into something other than a vector space. :)

(I didn't just mess up horribly, right? xD)

 

[kq6up]

Thanks, I was getting confuses by the fact that they refer to 1 as a polynomial, but I guess it is.

So that requirement that f(1) = 1 turns it into something other than a vector space. :)

You mean that it turns it into something that (by definition) is not part of the set we have already defined, but f(x) was g(x) a part of that set. Since g+f=2 our definition fails because it does not satisfy the conditions of a proper vector -- that is, Closure.

I think it is starting to slowly soak in. I am not used to being this knee deep in abstract math. Everything I have dealt with thus far has been very physical, and intuitive.

Thanks,
Chris Maness

 

[ModestyKing]

Thanks, I was getting confuses by the fact that they refer to 1 as a polynomial, but I guess it is.

You mean that it turns it into something that (by definition) is not part of the set we have already defined, but f(x) was g(x) a part of that set. Since g+f=2 our definition fails because it does not satisfy the conditions of a proper vector -- that is, Closure.

Hi Chris,

Well, 1 is a polynomial, but what they're saying is that they're putting a restriction on the polynomials. ALL polynomials in this vector space must be 1 when evaluated at one. By definition of a vector space, you must be able to add two of the vectors (elements of the set) and get a third element of the set. But if you add two polynomials who evaluate to 1, that is 1 + 1 = 2.

So yes, it doesn't satisfy closure. :)

 

[Matterwave]

Thanks, I was getting confuses by the fact that they refer to 1 as a polynomial, but I guess it is.



You mean that it turns it into something that (by definition) is not part of the set we have already defined, but f(x) was g(x) a part of that set. Since g+f=2 our definition fails because it does not satisfy the conditions of a proper vector -- that is, Closure.

I think it is starting to slowly soak in. I am not used to being this knee deep in abstract math. Everything I have dealt with thus far has been very physical, and intuitive.

Thanks,
Chris Maness

I think you might have a slight confusion. f(x) and g(x) are not equal to 1. f(1)=1 and g(1)=1, but f(x) might equal for example f(x)=x^3, and g(x) might be g(x)=x^2. Both satisfy f(1)=g(1)=1. When you add them together you get h(x)=f(x)+g(x)=x^3+x^2 which is still a polynomial of degree 3, but which no longer satisfies h(1)=1 as now h(1)=2.

 

[kq6up]

Yes it is finally starting to soak in. I am also reading a GPL (Free) Linear algebra textbook that is opening this up for me too. So I am correct in saying V:={f(1),g(1),1} (where f(1)=g(1)=1) is all that is a part of that failed vector space? So maybe if I defined V like V:={f(1),g(1),1,2} it would now have closure since 2 is now part of the set.

Thanks,
Chris Maness

 

[Matterwave]

What set are you defining?

The set you defined earlier was V={f(x)|f(1)=1 and f(x) is a polynomial of degree 3 or less}

This is read "the set V is the set of all function f(x) such that f(1) is equal to 1 and f(x) is a polynomial of degree 3 or less".

I don't understand what you mean with your notation.

If you meant V={f(x)|f(1)=1 or 2 and f(x) is a polynomial of degree 3 or less}, then no this new set doesn't form a vector space either. You will have, in this set elements where f(1)=2. The sum of this element with any other element g(x) of the set will necessarily have h(1)=f(1)+g(1)>2, and now h(x) is not in your set.

 

[kq6up]

I thought of a example. Say my set is V:={i,j,k} where i,j,k are unit vectors. I need to allow something akin to the cross product to be one of the operations so that I get the other vectors as a product (in the general since) of two of the vectors (e.g. ixj=k). I think I might be missing 0, and +/- in front as well to meet all 10 properties of a proper vector space. However, what I have shown is the property of closure. Correct?

Thanks,
Chris Maness

 

[kq6up]

What set are you defining?

The set you defined earlier was V={f(x)|f(1)=1 and f(x) is a polynomial of degree 3 or less}

This is read "the set V is the set of all function f(x) such that f(1) is equal to 1 and f(x) is a polynomial of degree 3 or less".

I don't understand what you mean with your notation.

If you meant V={f(x)|f(1)=1 or 2 and f(x) is a polynomial of degree 3 or less}, then no this new set doesn't form a vector space either. You will have, in this set elements where f(1)=2. The sum of this element with any other element g(x) of the set will necessarily have h(1)=f(1)+g(1)>2, and now h(x) is not in your set.

We were posting at the same time, so take the my last post as occurring before yours.

If we were going to add g(1)=1 to f(1)=1, would not g(1)=1 need to be a part of the original set?

Sorry, if my notation seems confusing or ill formed. I have never worked on math that is this abstract before, so I am an admitted hack.

Thanks,
Chris Maness

 

[Matterwave]

I thought of a example. Say my set is V:={i,j,k} where i,j,k are unit vectors. I need to allow something akin to the cross product to be one of the operations so that I get the other vectors as a product (in the general since) of two of the vectors (e.g. ixj=k). I think I might be missing 0, and +/- in front as well to meet all 10 properties of a proper vector space. However, what I have shown is the property of closure. Correct?

Thanks,
Chris Maness

Are you trying to make the cross product the "addition of vectors" operation? You can't do this because addition must be commutative, while the cross product is not.

[STRIKE]Your set is closed under the operation of cross product. That's about all you can say about your set.[/STRIKE] It's not even a group under the cross product operation since there is no identity element. It doesn't even come close to defining a vector space.

EDIT: Actually, I just realized your set is not even closed under the cross product as jxi=-k, which is not in your set.

 

[Matterwave]

We were posting at the same time, so take the my last post as occurring before yours.

If we were going to add g(1)=1 to f(1)=1, would not g(1)=1 need to be a part of the original set?

Sorry, if my notation seems confusing or ill formed. I have never worked on math that is this abstract before, so I am an admitted hack.

Thanks,
Chris Maness

I think you have some deep misunderstanding, but it's hard for me to figure out what, because it's hard for me to decipher your language.

g(1)=1 is not an element, it's an equation. g(x) (the whole function, not it's value at some x) is an element. If g(x) satisfies the equation g(1)=1, then g(x) is in the set you originally identified. If not, then g(x) is not in the set.

 

[kq6up]

I think you have some deep misunderstanding, but it's hard for me to figure out what, because it's hard for me to decipher your language.

g(1)=1 is not an element, it's an equation. g(x) (the whole function, not it's value at some x) is an element. If g(x) satisfies the equation g(1)=1, then g(x) is in the set you originally identified. If not, then g(x) is not in the set.

Yes, that helps. g(1) is part of the set because for our purposes it is identical to f(1).

Returning to my little thought experiment.

1. Expand V to include {+/-i,+/-j,+/-k,0,1} where dividing and taking the modulus of any vector is allowed.

Am I getting closer? If not, how is i,j,k properly defined as a vector space? It seems obvious that it is.

Edit: I added a line above after "Yes, that helps"

Thanks,
Chris Maness

 

[Matterwave]

Yes, that helps. g(1) is part of the set because for our purposes it is identical to f(1).

Really, I would NEVER say g(1) is part of the set (although technically it is since g(1)=1 and 1 is in the set as a polynomial of degree 0), this is seriously confusing language. I would always say g(x) is part of the set if g(1)=1.

Returning to my little thought experiment.

1. Expand V to include {+/-i,+/-j,+/-k,0,1} where dividing and taking the modulus of any vector is allowed.

Am I getting closer? If not, how is i,j,k properly defined as a vector space? It seems obvious that it is.

Edit: I added a line above after "Yes, that helps"

Thanks,
Chris Maness

i,j,k do NOT define a vector space, they are the basis of a vector space. The vector space for which they are the basis for is $\mathbb{R}^3$.

The vector space $\mathbb{R}^3$ consists of not only i,j,k (which is only 3 elements!) but of ALL sums of i,j,k over the real numbers. This means that 3i+2j+4k is in the set, and so is .1i+3.23j-4.3k. In general, all vectors of the form $\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ are in the set.

There is an uncountably infinite number of elements in this vector space. The elements are closed under vector addition obviously, and they are closed under vector additions over the field of real numbers. Vector addition is commutative and associative. The negative of any element is also in the set (e.g. -i is the negative of i). The identity element is the 0 vector. And all the other properties of vector spaces are satisfied.

The cross product does NOT have anything to do with the vector space nature defined above. It is an ADDITIONAL structure that we added to the vector space.