# Math Help - Subspace of a polynomial

1. ## Subspace of a polynomial

I'm having a lot of trouble with this topic. I have no idea on what to do for these questions:

1) Show that the set $S=\{p\in\mathbb{P}_2(0)=1\}" alt="S=\{p\in\mathbb{P}_2(0)=1\}" /> is NOT a subspace of $\mathbb{P}_2$

The only thing I can think of doing for this question is making $p(x)=ax^2+bx+1$

2) Show that the set:

$S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" alt="S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" />

is a subspace of $\mathbb{P}_3$

Would I do the following:

$p(x)=ax^3+bx^2+cx+d$

$\Longrightarrow p'(x)=3ax^2+2bx+c$

$\Longrightarrow p''(x)=6ax+2b$

$p''(x)=0 \ \mbox{for} \ x=-\dfrac{b}{3a}$

2. Originally Posted by acevipa
I'm having a lot of trouble with this topic. I have no idea on what to do for these questions:

1) Show that the set $S=\{p\in\mathbb{P}_2(0)=1\}" alt="S=\{p\in\mathbb{P}_2(0)=1\}" /> is NOT a subspace of $\mathbb{P}_2$

The only thing I can think of doing for this question is making $p(x)=ax^2+bx+1$
You want to show that the given set is not a subspace of $\mathbb{P}_2$. Recall the definition of a vector subspace. You need to show that one of the vector subspace axioms does not hold in this case; that is, you need to give a counter example to one of the axioms.

2) Show that the set:

$S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" alt="S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" />

is a subspace of $\mathbb{P}_3$

Would I do the following:

$p(x)=ax^3+bx^2+cx+d$

$\Longrightarrow p'(x)=3ax^2+2bx+c$

$\Longrightarrow p''(x)=6ax+2b$

$p''(x)=0 \ \mbox{for} \ x=-\dfrac{b}{3a}$
No. You need to show that the given set is a subspace of $\mathbb{P}_3$. Again, recall the axioms of vector space and show that they hold for any polynomials in your set S.

I'll start with closure to addition: Let $p,q \in S$. Then $p''(x) = q''(x) = 0$ by definition of S. Now, note that $(p+q)'' = p'' + q'' = 0$, that is, $(p+q)'' = 0$ and so by definition of S, $(p+q) \in S$

Can you do the rest?

3. Originally Posted by Defunkt
You want to show that the given set is not a subspace of $\mathbb{P}_2$. Recall the definition of a vector subspace. You need to show that one of the vector subspace axioms does not hold in this case; that is, you need to give a counter example to one of the axioms.

No. You need to show that the given set is a subspace of $\mathbb{P}_3$. Again, recall the axioms of vector space and show that they hold for any polynomials in your set S.

I'll start with closure to addition: Let $p,q \in S$. Then $p''(x) = q''(x) = 0$ by definition of S. Now, note that $(p+q)'' = p'' + q'' = 0$, that is, $(p+q)'' = 0$ and so by definition of S, $(p+q) \in S$

Can you do the rest?
For the first question, do you because $p(0)=1$ do we say that the zero vector is not included and therefore the set is empty.

With the second question, how would you show that it is non-empty?

4. Originally Posted by acevipa
For the first question, do you because $p(0)=1$ do we say that the zero vector is not included and therefore the set is empty.
Since $p(0)=1$ for any $p \in S$ then the zero polynomial is not in S. This, of course, does not mean that S is empty. But one of the subspace axioms is that the space must have the identity element. Here it does not, therefore S is not a subspace.

With the second question, how would you show that it is non-empty?
Can you think of any polynomial $p \in \mathbb{P}_3$ such that $p''(x) = 0 ~ \forall x \in \mathbb{R}$? (hint: its simple).

5. Originally Posted by Defunkt
Since $p(0)=1$ for any $p \in S$ then the zero polynomial is not in S. This, of course, does not mean that S is empty. But one of the subspace axioms is that the space must have the identity element. Here it does not, therefore S is not a subspace.

Can you think of any polynomial $p \in \mathbb{P}_3$ such that $p''(x) = 0 ~ \forall x \in \mathbb{R}$? (hint: its simple).
Okay, so would I just prove that a degree one polynomial $p(x)=ax+b$ is a subspace of $\mathbb{P}_3$?