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About Subspace problems

chukkitty

New member
May 15, 2013
2
Hello,

I want ask some subspace problems. Attachment is a question.
contains all polynomials with degree less than 3 and with real coefficients.

I want prove that item 1 and item 2 are subspace or not.

Am I insert real number to the item 1& 2 equation to test as follows:
(a) 0 ∈ S.
(b) S is closed under vector addition.
(c) S is closed under scalar multiplication.

Would you mind tell me how Determine whether the following sets are subspaces of P3. If it is a subspace, prove it. If it is not a subspace, give a counter-example.

Best Regard
Kitty
 

Attachments

ModusPonens

Well-known member
Jun 26, 2012
45
Indeed what you have to do is to provide a proof that the sets have the 0, are closed to addition and scalar multiplication.

The important thing, that works for all problems of this type is that, when you have a set $\{ x: C(x) \}$, where $C(x)$ is the condition for the x to belong to the set, to verify that:

1-C(0) is true. This implies that 0 is in the set;
2-If C(x) and C(y) is true (equivalently, if x and y are in the set), then C(x+y) is true. This implies that x+y is in the set;
3- If C(x) is true, then $C(ax)$ is true, where $a$ is a scalar. This implies that $ax$ is in the set.


The condition C for the first set in your exercise is a condition on a polinomial p. $C(p)$ is $p=a+bx+cx^2 \wedge a,b,c\in \mathbb{Q}$.

Teach a man how to fish... ;)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
(a) 0 ∈ S.
(b) S is closed under vector addition.
(c) S is closed under scalar multiplication.
In general the first is not needed since it is already covered in the third condition . Since the set of rational numbers is closed under addition and multiplications , item 1 forms a subspace .

For item 2 , I think there is a mistake since you are not specifying what is c ?
 

ModusPonens

Well-known member
Jun 26, 2012
45
In general the first is not needed since it is already covered in the third condition . Since the set of rational numbers is closed under addition and multiplications , item 1 forms a subspace .

For item 2 , I think there is a mistake since you are not specifying what is c ?
I think item 1 is not a subspace because it is not closed by multiplication by real scalars.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
In general the first is not needed since it is already covered in the third condition
Not always. Previously we need to prove that $S\neq \emptyset$ (so, the best is to prove $0\in S$).
 

chukkitty

New member
May 15, 2013
2
Dear All,

Thank you for your assistance.

I would try to prove two items to belong subspace or not. Because I don't know to write some Maths symbol on internet. So please refer to attachment of my work image.

(ITEM 1) The set contains the zero vector, as a=b=c=0 given p(x)=0+0x+0x².
Let p₁(x)=a₁+b₁x+c₁x² and p₂(x)=a₂+b₂x+c₂x².
Then p₁(x)+p₂(x)=(a₁+a₂)+(b₁+b₂)x+(c₁+c₂)x²
This polynomial has the correct form for a vector in U, so U is closed under vector addition.
Let p(x)=a+bx+cx² and α=ℝ and a,b,c∈ℚ.
Then αp(x)=(αa)+(αb)x+(αc)x²
This vector does not have the correct form for a vector in U if α can be irrational number, so U is not closed under scalar multiplication.
A counter-example: p(x)=(1/7)+(2/(11))x+(5/3)x²
Let α=√(13), then αp(x)=((√(13))/7)+((2√(13))/(11))x+((5√(13))/3)x². Show that the coefficients of x⁰,x¹,x² are not rational number.


(ITEM 2) The set contains the zero vector, as b=0 and any c given p(x)=0x+0x².
Let p₁(x)=b₁x+cx² and p₂(x)=b₂x+cx²
p₁(x)+p₂(x)= (b₁+b₂)x+(2c)x², where any c
This polynomial has the correct form for a vector in P₂, so P₂ is closed under vector addition.

Let p(x)=bx+cx² and α=ℝ
Then αp(x)=(αb)x+(αc)x² , where any c
This polynomial has the correct form for a vector in P₂, so P₂ is closed under scalar multiplication.
So P₂ is a subspace of P₃.

Please give me advise.

Thank you very much!

Kitty
 

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