Verifying Subspace of P3: Closure of Addition & Scalar Multiplication

[Sho Kano]

Homework Statement

Determine if the following is a subspace of $P_3$.
All polynomials $a_0+a_1x+a_2x^2+a_3x^3$ for which $a_0+a_1+a_2+a_3=0$

Homework Equations

use closure of addition and scalar multiplication

The Attempt at a Solution

Let $P=a_0+a_1x+a_2x^2+a_3x^3$ and $Q=b_0+b_1x+b_2x^2+b_3x^3$
$P+Q=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+(a_3+b_3)x^3$
am I missing a step? how can I show that matches the conditions?

 

[fresh_42]

Homework Statement

Determine if the following is a subspace of $P_3$.
All polynomials $a_0+a_1x+a_2x^2+a_3x^3$ for which $a_0+a_1+a_2+a_3=0$

Homework Equations

use closure of addition and scalar multiplication

The Attempt at a Solution

Let $P=a_0+a_1x+a_2x^2+a_3x^3$ and $Q=b_0+b_1x+b_2x^2+b_3x^3$
$P+Q=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+(a_3+b_3)x^3$
am I missing a step?

Yes, because multiples $c \cdot P$ for whatever $c$ is from, must also be a polynomial that satisfies the condition.

how can I show that matches the conditions?

What is the condition? Does $P+Q$ satisfies it?

 

[member 587159]

Homework Statement

Determine if the following is a subspace of $P_3$.
All polynomials $a_0+a_1x+a_2x^2+a_3x^3$ for which $a_0+a_1+a_2+a_3=0$

Homework Equations

use closure of addition and scalar multiplication

The Attempt at a Solution

Let $P=a_0+a_1x+a_2x^2+a_3x^3$ and $Q=b_0+b_1x+b_2x^2+b_3x^3$
$P+Q=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+(a_3+b_3)x^3$
am I missing a step? how can I show that matches the conditions?

You should also check whether the set is non-empty, which obviously is the case (why?).

 

[Sho Kano]

You should also check whether the set is non-empty, which obviously is the case (why?).

the set is non-empty because all of the coefficients don't necessarily have to be 0?

 

[Sho Kano]

What is the condition? Does P+QP+QP+Q satisfies it?

I think it does, is this right? $a_{ 0 }+a_{ 1 }+a_{ 2 }+a_{ 3 }=0\\ b_{ 0 }+b_{ 1 }+b_{ 2 }+b_{ 3 }=0\\ a_{ 0 }+a_{ 1 }+a_{ 2 }+a_{ 3 }+b_{ 0 }+b_{ 1 }+b_{ 2 }+b_{ 3 }=0+0=0\\ (a_{ 0 }+b_{ 0 })+(a_{ 1 }+b_{ 1 })+(a_{ 2 }+b_{ 2 })+(a_{ 3 }+b_{ 3 })=0$

 

[fresh_42]

the set is non-empty because all of the coefficients don't necessarily have to be 0?

No, empty set means no elements. But the polynomial $P=0=0+0x+0x^2+0x^3$ is not only an element, it is even needed, because a vector space must have $0$ in it.

I think it does, is this right? $a_{ 0 }+a_{ 1 }+a_{ 2 }+a_{ 3 }=0\\ b_{ 0 }+b_{ 1 }+b_{ 2 }+b_{ 3 }=0\\ a_{ 0 }+a_{ 1 }+a_{ 2 }+a_{ 3 }+b_{ 0 }+b_{ 1 }+b_{ 2 }+b_{ 3 }=0+0=0\\ (a_{ 0 }+b_{ 0 })+(a_{ 1 }+b_{ 1 })+(a_{ 2 }+b_{ 2 })+(a_{ 3 }+b_{ 3 })=0$

Yes. And next also for $c\cdot P$ with a scalar $c$.

 

[Mark44]

I think it does, is this right? $a_{ 0 }+a_{ 1 }+a_{ 2 }+a_{ 3 }=0\\ b_{ 0 }+b_{ 1 }+b_{ 2 }+b_{ 3 }=0\\ a_{ 0 }+a_{ 1 }+a_{ 2 }+a_{ 3 }+b_{ 0 }+b_{ 1 }+b_{ 2 }+b_{ 3 }=0+0=0\\ (a_{ 0 }+b_{ 0 })+(a_{ 1 }+b_{ 1 })+(a_{ 2 }+b_{ 2 })+(a_{ 3 }+b_{ 3 })=0$

As is, this doesn't make much sense. Start by assuming you have two functions that belong to your set. 1) Show that their sum is in the set. 2) Show that any scalar multiple of either of them is in the set.

 

[Mark44]

No, empty set means no elements. But the polynomial $P=0=0+0x+0x^2+0x^3$ is not only an element, it is even needed, because a vector space must have $0$ in it.

Of course that polynomial is in the given vector space, but it wouldn't hurt to show that it is also in the subset of P₃ defined by the equation $a_1 + a_2 + a_3 + a_4 = 0$.

 

[Sho Kano]

Yes. And next also for c⋅Pc⋅Pc\cdot P with a scalar ccc.

$k\in \Re \\ kP=ka_{ 0 }+ka_{ 1 }x+ka_{ 2 }x^{ 2 }+ka_{ 3 }x^{ 3 }\\ ka_{ 0 }+ka_{ 1 }+ka_{ 2 }+ka_{ 3 }=k(...)=0$
so the subset is a subspace?

 

[Sho Kano]

As is, this doesn't make much sense. Start by assuming you have two functions that belong to your set. 1) Show that their sum is in the set. 2) Show that any scalar multiple of either of them is in the set.

sorry I don't know if I wrote the set notation right
$S\in \left\{ a_{ 0 }+a_{ 1 }x+a_{ 2 }x^{ 2 }+a_{ 3 }x^{ 3 }|a_{ 0 }+a_{ 1 }+a_{ 2 }+a_{ 3 }=0 \right\} \\ P,Q\in S\\ P+Q=(a_{ 0 }+b_{ 0 })+(a_{ 1 }+b_{ 1 })x+(a_{ 2 }+b_{ 2 })x^{ 2 }+(a_{ 3 }+b_{ 3 })x^{ 3 }\\ (a_{ 0 }+b_{ 0 })+(a_{ 1 }+b_{ 1 })+(a_{ 2 }+b_{ 2 })+(a_{ 3 }+b_{ 3 })=0\\ k\in \Re \\ kP=ka_{ 0 }+ka_{ 1 }x+ka_{ 2 }x^{ 2 }+ka_{ 3 }x^{ 3 }\\ ka_{ 0 }+ka_{ 1 }+ka_{ 2 }+ka_{ 3 }=k(...)=0$

 

[fresh_42]

Yes. You should only replace the first $"\in "$ by $"="$.

Edit: Do you know why these two imply $0 \in S$ and $-P \in S$ for $P \in S\,$?
They are needed, because the vector space has to be a (additive) group. Why don't you have to show this separately?

 

[member 587159]

the set is non-empty because all of the coefficients don't necessarily have to be 0?

No that's not the right explanation. Look at @fresh_42 comment for the right answer. You should always check that the set is non-empty, because only then you can show it is a subspace by showing the set is closed under vector addition and scalar multiplication. The easiest way to do this is by checking whether the zero vector is in it (which has to be part of every vector space). However, you can also take another vector and show that this vector is in the set.

 

[pasmith]

Homework Statement

Determine if the following is a subspace of $P_3$.
All polynomials $a_0+a_1x+a_2x^2+a_3x^3$ for which $a_0+a_1+a_2+a_3=0$

Homework Equations

use closure of addition and scalar multiplication

The Attempt at a Solution

Let $P=a_0+a_1x+a_2x^2+a_3x^3$ and $Q=b_0+b_1x+b_2x^2+b_3x^3$
$P+Q=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+(a_3+b_3)x^3$
am I missing a step? how can I show that matches the conditions?

The condition on p : x \mapsto a_0 + a_1x + a_2x^2 + a_3x^3 is equivalent to p(1) = 0. What is cp(1) + q(1) if p(1) = q(1) = 0 and c \in \mathbb{R}?