# Linear subspaces and dimensions Proof

1. Sep 26, 2014

### Cassi

1. The problem statement, all variables and given/known data
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)

2. Relevant 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.

3. The attempt at a solution
I
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.

Last edited: Sep 26, 2014
2. Sep 26, 2014

### vela

Staff Emeritus
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. Sep 26, 2014

### LCKurtz

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.

4. Sep 26, 2014

### Cassi

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. Sep 26, 2014

### LCKurtz

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.

6. Sep 26, 2014

### Cassi

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. Sep 26, 2014

### LCKurtz

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. Sep 26, 2014

### Cassi

Thank you, this helps. I knew what the conclusion needed to be but had a hard time phrasing it. Thank you!

9. Sep 27, 2014

### Fredrik

Staff Emeritus
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.