Linear subspaces and dimensions Proof

In summary: S.I'm sorry I forgot to write the given condition. I just edited the post to add it. Yes, the condition if f(0)=f(1). I am very confused as how to prove all of this. I am trying to prove that S is a subset of Pn and prove the 2 closure axioms for S.
  • #1
Cassi
18
0

Homework Statement


Let Pn denote the linear space of all real polynomials of degree </= n, where n is fixed. Let S denote the set of all polynomials f in Pn satisfying the condition given. Determine whether or not S is a subspace of Pn. If S is a subspace, compute dim S.

The given condition if f(0)=f(1)

Homework Equations


Closure axioms:
(1)Closure under addition: For every pair of elements x and y in V there corresponds a unique element in V called the sum of x and y denoted by x+y.
(2) Closure under multiplication: For every x and y in V and every real number a there corresponds an element in V called the product of a and x, denoted by ax.

The Attempt at a Solution


I[/B] know that S is a subspace of Pn if S is a subset of Pn and it satisfies the closure axioms. So I need to prove these three things, but they seem trivial to me.

S c Pn because S: f(0)=f(1) has all elements in Pn. Is this enough of a proof to draw the conclusion?

For the closure axioms, I know that I have to show that f(x)+g(x) is an element in Pn, but again this seems trivial, I am unsure how to do this. Same for the closure axiom of multiplication.
 
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  • #2
I take is S is defined as ##S = \{ f \in P_n\ \vert\ f(0)=f(1) \}##. If so, why would you think every element of ##P_n## is in S? What about f(x)=x? It's in ##P_n## (if n>0) but ##f(0) \ne f(1)##.

Or is your reasoning that all elements of S are in ##P_n##, so closure is automatically satisfied? It's true that closure is satisfied for ##P_n##, e.g., f+g will be in ##P_n##, but you can't say the same thing about S, i.e., f+g isn't necessarily in S even if f and g are both in S.
 
  • #3
Cassi said:

Homework Statement


Let Pn denote the linear space of all really polynomials of degree </= n, where n is fixed. Let S denote the set of all polynomials f in Pn satisfying the condition given. Determine whether or not S is a subspace of Pn. If S is a subspace, compute dim S.

Is it a secret or are you going to tell us what the "condition given" is?

Homework Equations


Closure axioms:
(1)Closure under addition: For every pair of elements x and y in V there corresponds a unique element in V called the sum of x and y denoted by x+y.
(2) Closure under multiplication: For every x and y in V and every real number a there corresponds an element in V called the product of a and x, denoted by ax.

The Attempt at a Solution


I[/B] know that S is a subspace of Pn if S is a subset of Pn and it satisfies the closure axioms. So I need to prove these three things, but they seem trivial to me.

S c Pn because S: f(0)=f(1) has all elements in Pn. Is this enough of a proof to draw the conclusion?

Is f(0) = f(1) the secret condition?

For the closure axioms, I know that I have to show that f(x)+g(x) is an element in Pn, but again this seems trivial, I am unsure how to do this. Same for the closure axiom of multiplication.

You haven't shown closure, If ##f,g\in S## what does that give you? Then show ##cf## and ##f+g## are in ##S## too.
 
  • #4
LCKurtz said:
Is it a secret or are you going to tell us what the "condition given" is?
Is f(0) = f(1) the secret condition?
You haven't shown closure, If ##f,g\in S## what does that give you? Then show ##cf## and ##f+g## are in ##S## too.
I'm sorry I forgot to write the given condition. I just edited the post to add it. Yes, the condition if f(0)=f(1). I am very confused as how to prove all of this. I am trying to prove that S is a subset of Pn and prove the 2 closure axioms for S.
 
  • #5
Cassi said:
I'm sorry I forgot to write the given condition. I just edited the post to add it. Yes, the condition if f(0)=f(1). I am very confused as how to prove all of this. I am trying to prove that S is a subset of Pn and prove the 2 closure axioms for S.
Cassi said:
I'm sorry I forgot to write the given condition. I just edited the post to add it. Yes, the condition if f(0)=f(1). I am very confused as how to prove all of this. I am trying to prove that S is a subset of Pn

Isn't ##S## given to be a subset satisfying f(0)=f(1)? You don't have to prove it is a subset.

and prove the 2 closure axioms for S.

In post #4 at the end I asked:

If f,g∈S what does that give you? Once you answer that you can talk about whether cf and f+g are in S.
 
  • #6
LCKurtz said:
Isn't ##S## given to be a subset satisfying f(0)=f(1)? You don't have to prove it is a subset.
In post #4 at the end I asked:

If f,g∈S what does that give you? Once you answer that you can talk about whether cf and f+g are in S.
So would I say, let f and g be in S such that they satisfy the given condition. Now f+g exist in S because f(0)+g(0)=f(1)+g(1) therefore the first axiom stands. And for the second cf exists in S because cf(0)=cf(1), therefore the second axiom stands. Is this a valid argument?
 
  • #7
Cassi said:
So would I say, let f and g be in S such that they satisfy the given condition. Now f+g exist in S because f(0)+g(0)=f(1)+g(1) therefore the first axiom stands. And for the second cf exists in S because cf(0)=cf(1), therefore the second axiom stands. Is this a valid argument?

That's the idea, but you could certainly phrase it better. Instead of saying "let f and g be in S such that they satisfy the given condition" you could say: Suppose ##f,g\in S##. Then f(0)=f(1) and g(0) = g(1). Adding these equations gives f(0)+g(0) = f(1)+g(1) so ##f+g\in S##. It is very simple but needs to be stated properly. Similarly for the ##cf## case.
 
  • #8
LCKurtz said:
That's the idea, but you could certainly phrase it better. Instead of saying "let f and g be in S such that they satisfy the given condition" you could say: Suppose ##f,g\in S##. Then f(0)=f(1) and g(0) = g(1). Adding these equations gives f(0)+g(0) = f(1)+g(1) so ##f+g\in S##. It is very simple but needs to be stated properly. Similarly for the ##cf## case.
Thank you, this helps. I knew what the conclusion needed to be but had a hard time phrasing it. Thank you!
 
  • #9
Cassi said:
let f and g be in S such that they satisfy the given condition.
Seems like a strange thing to say when all elements of S satisfy the given condition.

This is how I usually phrase these things: Let ##f,g\in S## be arbitrary. We have
$$(f+g)(0)=f(0)+g(0)=f(1)+g(1)=(f+g)(1).$$ This implies that ##f+g\in S##.

This version makes it very clear that we're using the definition of f+g. (If we don't use it, we can't claim to have proved a statement about f+g, so it's pretty essential to make it clear that we're using it). Similarly, you should make it clear that you're using the definition of cf when you prove the other one.
 

Related to Linear subspaces and dimensions Proof

1. What is a linear subspace?

A linear subspace is a subset of a vector space that satisfies the properties of a vector space. This means that it is closed under addition and scalar multiplication.

2. What is the dimension of a linear subspace?

The dimension of a linear subspace is the number of linearly independent vectors that span the subspace. It is also equal to the number of coordinates needed to uniquely describe any vector in the subspace.

3. How do you prove that a set of vectors form a linear subspace?

To prove that a set of vectors form a linear subspace, you need to show that they satisfy the properties of a vector space. This means that they are closed under addition and scalar multiplication, and that they contain the zero vector.

4. How do you find the basis of a linear subspace?

To find the basis of a linear subspace, you need to find a set of linearly independent vectors that span the subspace. This can be done by using Gaussian elimination or by finding the null space of the subspace's matrix representation.

5. Can a linear subspace have a dimension greater than the dimension of its vector space?

No, the dimension of a linear subspace cannot be greater than the dimension of its vector space. This is because the subspace is a subset of the vector space and therefore it cannot have more linearly independent vectors than the vector space itself.

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