# Homework Help: Polynomial as a subspace?

1. Feb 17, 2010

### Susanne217

Hi

I am presented with the following problem

1. The problem statement, all variables and given/known data

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 $$P_4(\mathbb{R})$$
2. Relevant 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
2) closed under scalar multiplications
3. 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.

Last edited: Feb 17, 2010
2. Feb 17, 2010

### 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?

3. Feb 17, 2010

### Susanne217

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?

4. Feb 17, 2010

### 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.

5. Feb 17, 2010

### rsa58

the coefficients aren't zero, they SUM to zero

6. Feb 17, 2010

### 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.

7. Feb 17, 2010

### rsa58

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

8. Feb 17, 2010

### Susanne217

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?

9. Feb 17, 2010

### Staff: Mentor

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.

10. Feb 17, 2010

### 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.)

11. Feb 17, 2010

### 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.

12. Feb 17, 2010

### 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.

13. Feb 17, 2010

### Staff: Mentor

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

14. Feb 17, 2010

### Susanne217

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?

15. Feb 17, 2010

### vela

Staff Emeritus
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.

16. Feb 17, 2010

### Susanne217

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?

17. Feb 17, 2010

### vela

Staff Emeritus
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.

18. Feb 17, 2010

### Susanne217

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?

19. Feb 17, 2010

### vela

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

20. Feb 17, 2010

### Susanne217

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?

Last edited: Feb 17, 2010