Is S a Subspace of P4?

[Susanne217]

Hi

I am presented with the following problem

Homework Statement

Let $$P_{4}(\mathbb{R}) = \{a_0 + a_1 \cdot x + a_2 \cdot x^2 + a_3 \cdot x^3|a_0,a_1,a_2,a_3 \in \mathbb{R} \}$$

Then let S be a subset of $$P_4(\mathbb{R})$$ where

$$S = \{a_0 + a_1 \cdot x + a_2 \cdot x^2 + a_3 \cdot x^3|a_0+a_1+a_2+a_3 = 0 \}$$
Show that S is a subspace of $$P_4(\mathbb{R})$$

Homework Equations

I that according to theory that for a space to be a subspace of a vector space then it has to comply with two axioms
1) closed under addition
2) closed under scalar multiplications

The Attempt at a Solution

Since $$since (x,x^2,x^3)$$ belongs to S then $$\{a_0 + a_1 \cdot x + a_2 \cdot x^2 + a_3 \cdot x^3 \}$$ also belongs to S and thusly condition 1) holds?

Since (x,x^2,x^3) belongs to S then as since $$a_0,a_1,a_2,a_3$$ belongs to $$\mathbb{R}$$ then $$\{a_0 + a_1 \cdot x + a_2 \cdot x^2 + a_3 \cdot x^3 \}$$ also belongs to S. Thus condition 2) Holds and hence is a subspace of the original space.

 

[rsa58]

does zero belong to the subset? do any two vectors from the set belong again tothe set? in other words when you add two polynomials whose coefficients sum to 0, does the resulting polynomial's coefficients sum to 0? can you figure out what happens when you multiply by a scalar? is c*0=0?

 

[Susanne217]

does zero belong to the subset? do any two vectors from the set belong again tothe set? in other words when you add two polynomials whose coefficients sum to 0, does the resulting polynomial's coefficients sum to 0? can you figure out what happens when you multiply by a scalar? is c*0=0?

As I see it since the coefficients are zero then S sums to the zero polynomial and then zero vector belongs to S. Isn't that what you mean?

 

[rsa58]

an element in S looks like the first part of the bracketed expression. for example 0 is in S because 0 = 0 +0*x + 0*x^2 ...+0*x^4 where the coefficients sum to zero.

 

[rsa58]

the coefficients aren't zero, they SUM to zero

 

[rsa58]

so (x,x^2,x^3) does not belong to S. also this notation is vague don't use it. do you mean each of these vectors belongs to S? if so this is incorrect the coefficients of the monomials are not zero. do you mean the vector x + x^2 +x^3 belongs to S? if so incorrect the coefficients don't sum to zero.

 

[rsa58]

sorry 0=0+0*x + 0*x^2 +0*x^3 not what i wrote

 

[Susanne217]

so (x,x^2,x^3) does not belong to S. also this notation is vague don't use it. do you mean each of these vectors belongs to S? if so this is incorrect the coefficients of the monomials are not zero. do you mean the vector x + x^2 +x^3 belongs to S? if so incorrect the coefficients don't sum to zero.

Hi RSA58,

Then I have misunderstood I thought that I had to view each component of the polynomial as vector components and then if they belong to S then their addition also belongs to S? And thus then S is closed under addition? I can see that if the coffients all are zero then the polynomial sum to the zero polynomial, but what I don't get does that make the whole of S to be closed under addition and is the answer that I am looking for?

 

[Mark44]

These polynomials belong to S:
1 - x² (= 1 + 0x - 1x² + 0x³)
0 + 0x + 0x² + 0x³
2 - x - x³

Every polynomial that belongs to S is such that its coefficients add up to zero.

These polynomials do not belong to S:
5 (= 5 + 0x + 0x² + 0x³)
(1/2)x²
1 + 2x + 3x² - 5x³
If the coefficients of a polynomial do not add up to zero, it doesn't belong to S.

Clearly the zero polynomial belongs to S. Now take two arbitrary polynomials in S and add them together. If their sum is also in S, the set is closed under addition.

Finally, take an arbitrary polynomial in S, and multiply it by a scalar. If the product is also in S, the set is closed under scalar multiplication. If the set includes the zero polynomial, is closed under addition, and is closed under scalar multiplication, set S is a subspace of P₄.

 

[HallsofIvy]

Here is what you need to show:
If u and v are in the set, then so is u+v

You want to show that if $a_0+ a_1x+ a_2x^2+ a_3 x^3$ and $b_0+ b_1x+ b_2x^2+ b_3x^3$ are in the set then so is $(a_0+b_0)+ (a_1+ b_1)x+ (a_2+ b_2)x^2+ (a_3+ b_3)x^3$.

In particular, you must show that if $a_0+ a_1+ a_2+ a_3= 0$ and $b_0+ b_1+ b_2+ b_3= 0$, then $(a_0+b_0)+ (a_1+b_1)+ (a_2+b_2)+ (a_3+ b_3)= 0$

You must also show that if u is in the set and s is a number, then su is in the set.

If $u= a_0+ a_1x+ a_2x^2+ a_3x^3$ then $su= sa_0+ sa_1x+ sa_2x^2+ sa_3x^3$. You must show that is $a_0+ a_1+ a_2+ a_3= 0$ then $sa_0+ sa_1+ sa_2+ sa_3= 0$

Finally, you have already observed that $0+ 0x+ 0x^2+ 0x^3$ is the 0 vector and is in this set.

(Strictly speaking, you don't have to prove that the 0 vector is in the space. You do need to prove that the set is non-empty, that some vector is in the set and, usually, it is simplest to show that the 0 vector is in the set.)

 

[rsa58]

yeah, what you're saying makes sense to me but not the way you are writing it. you can write all polynomials in S as (a0,a1,a2,a3) where ai are from the field in this case R. we are not interested in the markers 1,x, ... but the coefficients. the vector is defined by the coefficients of the polynomial and its degree.

 

[rsa58]

ah because if the set is non empty and closed under scalar multiplication and addition then since every element has an additive inverse in the field then 0 belongs to the set. but in any case zero must be in the set because the set is also a vector space.

 

[Mark44]

ah because if the set is non empty and closed under scalar multiplication and addition then since every element has an additive inverse in the field then 0 belongs to the set. but in any case zero must be in the set because the set is also a vector space.

Well, the set is really a subspace of a vector space.

 

[Susanne217]

Here is what you need to show:
If u and v are in the set, then so is u+v

You want to show that if $a_0+ a_1x+ a_2x^2+ a_3 x^3$ and $b_0+ b_1x+ b_2x^2+ b_3x^3$ are in the set then so is $(a_0+b_0)+ (a_1+ b_1)x+ (a_2+ b_2)x^2+ (a_3+ b_3)x^3$.

In particular, you must show that if $a_0+ a_1+ a_2+ a_3= 0$ and $b_0+ b_1+ b_2+ b_3= 0$, then $(a_0+b_0)+ (a_1+b_1)+ (a_2+b_2)+ (a_3+ b_3)= 0$

You must also show that if u is in the set and s is a number, then su is in the set.

If $u= a_0+ a_1x+ a_2x^2+ a_3x^3$ then $su= sa_0+ sa_1x+ sa_2x^2+ sa_3x^3$. You must show that is $a_0+ a_1+ a_2+ a_3= 0$ then $sa_0+ sa_1+ sa_2+ sa_3= 0$

Finally, you have already observed that $0+ 0x+ 0x^2+ 0x^3$ is the 0 vector and is in this set.

(Strictly speaking, you don't have to prove that the 0 vector is in the space. You do need to prove that the set is non-empty, that some vector is in the set and, usually, it is simplest to show that the 0 vector is in the set.)

Hi HallsofIvy,
So what you are trying to lead me to conclude that since the zero vector clearly is in S, then both axioms the subspace apply to S?

 

[vela]

Hi HallsofIvy,
So what you are trying to lead me to conclude that since the zero vector clearly is in S, then both axioms the subspace apply to S?

No, HallsofIvy is saying you have to prove that S is closed under vector addition and scalar multiplication and that S is non-empty. It's possible to prove both closure conditions hold, but if S contains no vectors, it's not a subspace.

 

[Susanne217]

No, HallsofIvy is saying you have to prove that S is closed under vector addition and scalar multiplication and that S is non-empty. It's possible to prove both closure conditions hold, but if S contains no vectors, it's not a subspace.

Hi vela,

Maybe I am stupid or confused but one of the three main subspace definition isn't that to show that if a space contains the zero vector then its a vector subspace?

 

[vela]

I think what you mean may be getting lost in your wording, but it sounds to me like you're saying if the zero vector is in S, then it follows that the other two conditions automatically hold and therefore S is a subspace. That's not correct. To show S is a subspace, you have to show all three conditions hold. Showing that the zero vector is in S doesn't automatically guarantee the two closure requirements are true.

 

[Susanne217]

I think what you mean may be getting lost in your wording, but it sounds to me like you're saying if the zero vector is in S, then it follows that the other two conditions automatically hold and therefore S is a subspace. That's not correct. To show S is a subspace, you have to show all three conditions hold. Showing that the zero vector is in S doesn't automatically guarantee the two closure requirements are true.

Okay,

I get what you and Hallsoft is getting at but to show that points he/she is referings the only to test this isn't to choose arbitrary set of values of a and b's to test the points Hall is putting up?

 

[vela]

Okay,

I get what you and Hallsoft is getting at but to show that points he/she is referings the only to test this isn't to choose arbitrary set of values of a and b's to test the points Hall is putting up?

Sorry, I don't understand what you're asking here at all.

 

[Susanne217]

Sorry, I don't understand what you're asking here at all.

I mean isn't that only way to test the axioms of the subspace isn't to set e.g.

a0 = 1

a1 = -1

a2 = 0

a3 = 0

and then test the axioms on S?

btw. Why doesn't many of the Linear Algebra books deal with polynomials as vector spaces?

 

[vela]

No. I'm guessing your thinking is that if you show a requirement holds for certain specific vectors, that it'll follow that it is true for an arbitrary linear combination of those vectors. That approach won't work here. That would only prove that a requirement holds for that one specific choice of values for the $a_i$'s. You want to show it holds for all possible values of the $a_i$'s that satisfy $a_0+a_1+a_2+a_3=0$.

For example, to show S is closed under scalar multiplication, you want to show that if $p(x)\in S$ and $c \in R$, then $(cp)(x) \in S$. So let $p(x)=a_0+a_1 x+a_2 x^2 + a_3 x^3 \in S$. Because it's in S, you know that $a_0+a_1+a_2+a_3=0$, but there's no other restriction on the coefficients, so p(x) could be any element of S. Now you also have $(cp)(x) = (ca_0)+(ca_1)x+(ca_2)x^2+(ca_3)x^3$; to show that this function is in S, you need to show that its coefficients sum to 0. Well, that's true because

$$ca_0+ca_1+ca_2+ca_3=c(a_0+a_1+a_2+a_3)=c(0)=0$$.

Therefore, you can conclude that $(cp)(x)\in S$ and that S is closed under scalar multiplication.

You want to do the same sort of thing to show closure under vector addition.

 

[Mark44]

btw. Why doesn't many of the Linear Algebra books deal with polynomials as vector spaces?

Some do and some don't. The ones that do often call them function spaces, which have the same axioms as vector spaces.

 

[Susanne217]

Hi vela, Mark and Friends,

I understand you guys points and vela thanks for the example.

I did a search in google books and found a textbook on Linear Algebra by Larry Smith which says:

Remove the restriction $$P_4(\mathbb{R})$$ so that it becomes $$p(x) = a_0 + a_1\cdot x + a_2\cdot x^2 +a_3 \cdot x^3$$ and hence

vector = polynomial
vector addition = polynomial addition
scalar multiplication = multiplication of a polynomial by a number...

(I know that this is a repeat of what Mr/Miss Hallsofty wrote :)

So please correct and call me silly, but let me just imagine I take a polynomial of simular character as p(x) e.g. $$q(x) = b_0 + b_1\cdot x + b_2\cdot x^2 +b_3 \cdot x^3$$

with the condition that $$b_0+b_{1}+b_2+b_3 = 0$$

I then add the together which then gives

$$p(x)+q(x) = a_0+a_1 \cdot x + a_2\cdot x^2 +a_3 \cdot x^3 + b_0 + b_1\cdot x + b_2\cdot x^2 +b_3 \cdot x^3 = a_0 + b_0 + (a_{1}+b_{1}) \cdot x + (a_{2}+b_{2})\cdot x^2 + (a_3 + b_3) \cdot x^3$$

and since $$b_0+b_{1}+b_2+b_3 = 0$$ and $$a_0+a_{1}+a_2+a_3 = 0$$ then Our subset S must at most contain the zero vector and thus its non-empty. Cause $${a_0+\cdots+a_4} = 0$$ and Cause $${b_0+\cdots+b_4} = 0$$ the cause it doesn't matter if the a's are 0+1 - 1 + 0 and the b's are -1+1+2-2 etc they must always sum to zero and thus not only does the subspace contain the two vector (polynomials) p and q it also contain their product p+q and hence condition 1) or being a subspace is true. Am I onto something now :) ?

 

[Mark44]

Hi vela, Mark and Friends,

I understand you guys points and vela thanks for the example.

I did a search in google books and found a textbook on Linear Algebra by Larry Smith which says:

Remove the restriction $$P_4(\mathbb{R})$$ so that it becomes $$p(x) = a_0 + a_1\cdot x + a_2\cdot x^2 +a_3 \cdot x^3$$ and hence

vector = polynomial
vector addition = polynomial addition
scalar multiplication = multiplication of a polynomial by a number...

(I know that this is a repeat of what Mr/Miss Hallsofty wrote :)

So please correct and call me silly, but let me just imagine I take a polynomial of simular character as p(x) e.g. $$q(x) = b_0 + b_1\cdot x + b_2\cdot x^2 +b_3 \cdot x^3$$

with the condition that $$b_0+b_{1}+b_2+b_3 = 0$$

I then add the together which then gives

$$p(x)+q(x) = a_0+a_1 \cdot x + a_2\cdot x^2 +a_3 \cdot x^3 + b_0 + b_1\cdot x + b_2\cdot x^2 +b_3 \cdot x^3 = a_0 + b_0 + (a_{1}+b_{1}) \cdot x + (a_{2}+b_{2})\cdot x^2 + (a_3 + b_3) \cdot x^3$$

and since $$b_0+b_{1}+b_2+b_3 = 0$$ and $$a_0+a_{1}+a_2+a_3 = 0$$ then Our subset S must at most contain the zero vector and thus its non-empty. Cause $${a_0+\cdots+a_4} = 0$$ and Cause $${b_0+\cdots+b_4} = 0$$ the cause it doesn't matter if the a's are 0+1 - 1 + 0 and the b's are -1+1+2-2 etc they must always sum to zero and thus not only does the subspace contain the two vector (polynomials) p and q it also contain their product p+q and hence condition 1) or being a subspace is true. Am I onto something now :) ?

Sort of, but you're not saying it very well. You have shown that the set of functions (you haven't shown it's a subspace yet) contains the zero function, so the set contains at least (not at most) one function. Your work shows that if p(x) and q(x) are in the set, then so is p(x) + q(x), the sum of these functions (not their product). This means that the set is closed under addition. If you can show that c*p(x) is in the set whenever p(x) is in the set and c is an arbitrary real number, you will have shown that the set is closed under scalar multiplication. Then you can say that the set is actually a subspace of P₄.