# Determine whether the subset is a vector subspace

1. Jan 11, 2017

### BubblesAreUs

1. The problem statement, all variables and given/known data
Recall that F is the vector space of functions from ℝ to ℝ, with the usual operations of addition and scalar multiplication of functions. For each of the following subsets of F, write down two functions that belong to the subset, and determine whether or not the subset is a vector subspace of F.

a) The set of polynomials of degree equal to 3.

2. Relevant equations

Axions: Set ℙ ≠ ∅, ℙ is closed under addition and ℙ is closed under scalar multiplication.

3. The attempt at a solution

ℙ = { ƒ ∈ F | f(x) = ax3, a ∈ ℝ }

f(0) = a(0)3 = 0 Thus, ℙ is a non-empty set.

Let g, h ∈ ℙ then,

g(x) = bx3
h(x) = cx3

g + h = g(x) + h(x) = bx3 + cx3 = (b + c)x3

Thus g + h is closed under addition.

Let f ∈ ℙ Λ k ∈ ℝ then,

k⋅f = k⋅f(x) = k(ax3) = ak(x3) = (ak)x3

Thus k⋅f is closed under scalar multiplication.

Therefore ℙ is a vector space of F.

Last edited: Jan 11, 2017
2. Jan 11, 2017

### Staff: Mentor

You probably mean axioms, as axions are hypothetical particles in physics. But these aren't axioms either. It is the definition of a vector space. You don't need to require $\mathbb{P} \neq \emptyset$ separately, because a vector space always has a zero vector. This follows from the other properties. Do you know how?
There is a difference between a polynomial as a whole and its values at certain points, here $x=0$. For $f(x)=ax^3$ to be identically zero, it has to be zero for all possible values of $x$, not only for $x=0$. Thus is doesn't matter what $f(0)$ is. $f(x) = 0 \cdot x^3$ is the reason there is a zero and $\mathbb{P}$ isn't empty. Which brings us to the next point.
Of which degree is e.g. $f(x)= x^3 +2x^2+x+1\,$? And if $f(x)=0 \cdot x^3$ is in the vector space, which degree has it?
So how exactly is the degree of a polynomial defined? Is $\mathbb{P}$ then still a vector space?

3. Jan 11, 2017

### Staff: Mentor

That's axioms
Not quite. g + h isn't the set -- it's P. That set is closed under addition.
Again, P is closed under scalar multipilcation
I would say that P is a subspace of F.

4. Jan 11, 2017

### andrewkirk

That is not the set of polynomials of degree 3. The degree is the highest index of x in any term in the polynomial, so the general form of a polynomial of degree 3 is $ax^3+bx^2+cx+d$ where $a,b,c,d\in F$ and $a\neq 0$.
The second statement doesn't appear to follow from the first. Why not just write down a degree-3 polynomial to demonstrate non-emptiness? Choose the simplest one you can think of.

EDIT: I didn't see the above posts when I wrote this. As you were. Read them instead. They have covered all that's needed.

5. Jan 11, 2017

### Staff: Mentor

$\mathbb{R}\cdot x^3$ is a vector space. The set of all real polynomials of exactly degree $3$ is not.

6. Jan 11, 2017

### Staff: Mentor

Most textbooks I have seen will talk about the space of polynomials of degree less than or equal to some integer value, so I've been a little uncertain about what is meant in this problem; that is, if "exactly 3rd degree" is meant.

In any case, the set of polynomials of degree 3 is a subspace of the space of polynomials of degree <= 3, isn't it?
The set is certainly closed under function addition, as well as scalar multiplication. The only objection I can think of is whether 0x3 belongs to the set.

7. Jan 11, 2017

### Staff: Mentor

That's why I wrote $\mathbb{R}\cdot x^3$ because demanding a leading term of degree three isn't sufficient. $\mathbb{R}\cdot x^3$ solves the problem with zero (which made the OP's conclusions correct). But I stumbled upon this "equal to three" which has been the source of confusion here. One always has to take the polynomials up to degree $n$ to get it closed under addition. I thought I'd better resolve it, before it got worse. It seemed that all of us answered to a different question.

8. Jan 12, 2017

### Math_QED

About the empty part. I believe it is nessecary to show that the set is non empty. How else can he take 2 functions in the set and show that a linear combination is also part of the set?

EDIT: I figured you most likely meant that that shouldn't be part of the definition, and then I completely agree.

9. Jan 12, 2017

### Staff: Mentor

(Only a note at the margin:) I'm not completely sure, but I think $\emptyset$ is the basis of $\{0\}$.

10. Jan 12, 2017

### Math_QED

Yes that's true. But why does it matter? If the subspace would be the trivial null-vector space, we still have to show that there is an element in the subset (this would be the null vector). The empty set is not a vector space.

11. Jan 12, 2017

### Staff: Mentor

I haven't said it matters, therefore my remark as a note on the margin. I didn't and don't want to deepen this highly theoretical discussion here, because I don't think it helps the OP.

12. Jan 12, 2017

### Staff: Mentor

I haven't seen this notation before. What does it mean?

13. Jan 12, 2017

### Staff: Mentor

It means $\mathbb{R} \cdot x^3 = \{f\,\vert \,f=a\cdot x^3 \text{ for an } a \in \mathbb{R}\}$, i.e. the set of all real multiples of $x^3$ or the straight that is spanned by the vector $x^3$. The same as a one-dimensional real vector space with a basis $v$ can be written as $\mathbb{R}v$. It avoids the exception for $f=0$ which comes in, if we require the degree of $f$ to be $3$.