# Verifying subspaces

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1. Jan 28, 2017

### Sho Kano

1. The problem statement, all variables and given/known data
Determine if the following is a subspace of $P_3$.
All polynomials $a_0+a_1x+a_2x^2+a_3x^3$ for which $a_0+a_1+a_2+a_3=0$

2. Relevant equations
use closure of addition and scalar multiplication

3. 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?

2. Jan 28, 2017

### Staff: Mentor

Yes, because multiples $c \cdot P$ for whatever $c$ is from, must also be a polynomial that satisfies the condition.
What is the condition? Does $P+Q$ satisfies it?

3. Jan 28, 2017

### Math_QED

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

4. Jan 28, 2017

### Sho Kano

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

5. Jan 28, 2017

### Sho Kano

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

6. Jan 28, 2017

### Staff: Mentor

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

7. Jan 28, 2017

### Staff: Mentor

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.

8. Jan 28, 2017

### Staff: Mentor

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 $a_1 + a_2 + a_3 + a_4 = 0$.

9. Jan 28, 2017

### Sho Kano

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

10. Jan 28, 2017

### Sho Kano

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

11. Jan 28, 2017

### Staff: Mentor

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?

12. Jan 29, 2017

### Math_QED

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.

13. Jan 29, 2017

### pasmith

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 $c \in \mathbb{R}$?