# Verifying Subspace of P3: Closure of Addition & Scalar Multiplication

• Sho Kano
In summary: 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_{
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?

Sho Kano said:

## 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?

Sho Kano said:

## 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?).

Math_QED said:
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?

fresh_42 said:
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##

Sho Kano said:
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.
Sho Kano said:
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##.

Sho Kano said:
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.

fresh_42 said:
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##.

fresh_42 said:
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?

Mark44 said:
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##

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?

Sho Kano said:
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.

Sho Kano said:

## 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 $c \in \mathbb{R}$?

## 1. What is a subspace?

A subspace is a subset of a vector space that satisfies the three properties of closure under addition, closure under scalar multiplication, and containing the zero vector. In simpler terms, a subspace is a smaller vector space that is contained within a larger vector space.

## 2. How do you verify if a subset is a subspace of P3?

To verify if a subset is a subspace of P3 (the vector space of polynomials with degree less than or equal to 3), we need to check if it satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and containing the zero vector. This can be done by performing addition and scalar multiplication on the elements of the subset and checking if the results are also in the subset, and by ensuring that the zero polynomial (with all coefficients equal to 0) is included in the subset.

## 3. What is closure under addition?

Closure under addition is one of the three properties that a subset must satisfy to be considered a subspace. It means that when two elements of the subset are added together, the result must also be in the subset. In other words, the subset is closed under the operation of addition.

## 4. What is closure under scalar multiplication?

Closure under scalar multiplication is another property that a subset must satisfy to be considered a subspace. It means that when an element of the subset is multiplied by a scalar (a constant), the result must also be in the subset. In other words, the subset is closed under the operation of scalar multiplication.

## 5. Why is it important to verify if a subset is a subspace?

Verifying if a subset is a subspace is important because it ensures that the subset follows the same rules and properties as the larger vector space. This allows us to use the same operations and techniques on the subset as we would on the larger vector space. Additionally, knowing that a subset is a subspace can help us in solving problems and making calculations more efficiently.

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