Determine whether S is a subspace of $P_n$, the vectorspace of all real polynomial

Question

Determine whether S is a subspace of $P_n$, the vector space of all real polynomials of degree $\leq n-1 $ of this form:

\begin{equation} p = a_0 + a_1X+a_2X^2+ \dots + a_{n-1}X^{n-1} \end{equation}

To determine if the set $S = \left\{ p \in p_n(\mathbb{R}) \mid\forall \alpha \in \mathbb{R}: p(\alpha) = p(-\alpha) \right\}$ is a subspace, I need to check for these 3 things:

1. let q, p $\in S$, then $r = q+p \in S$
2. let $\mathbb{B} \in \mathbb{R}$ and $p \in S$, then $\mathbb{B}p \in S$
3. $\mathbf0 \in S$

So for number 1, we let $r(a) = q(a) + p(a)$ and because $q(a) + p(a) = q(-a) + p(-a)$ then $r(a) = r(-a)$.

And for number 2, we let $w(a) = \mathbb{B}p(a)$. and because $\mathbb{B}$ it's just a scaler $w(-a) = \mathbb{B}p(-a)$.

But how do I check for number 3, i.e showing that $\mathbf0 \in S$?

Answer 1

$\mathbf0$ is the zero polynomial, the null vector of the space $p_n(\mathbb{R})$, given with $\mathbf0(\alpha)=\sum_{i=0}^n0\cdot\alpha^i=0$ f.a. $\alpha$ as it works as the identity for pointwise addition of functions. Thus, especially $\mathbf0(\alpha)=0=\mathbf0(-\alpha)$ f.a. $\alpha\in\mathbb{R}$. Followingly, $\mathbf0\in S$.

Your work for the first two closure conditions look perfectly fine.

EDIT: A function $f$, i.e. in your case a polynomial of degree $n$(maximal), s.t. $f(x)=f(-x)$ is called an even function. Besides prominent examples like sine, especially all constant functions are even. (Can you see why?)

Answer 2

Other people have clarified to you what you are missing. I am presenting something that may be helpful to you in the future. After you learn about linear independence or linear maps, you can come back here and read this answer again.

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If you already have some knowledge about linear maps, then you can show that $S$ is an $\mathbb{R}$-vector subspace of $P_n$ of dimension $\left\lceil\dfrac{n}{2}\right\rceil$ by exhibiting a surjective linear map $\varphi:P_n\to\mathbb{R}^{\left\lfloor\frac{n}{2}\right\rfloor}$ and showing that $S=\ker(\varphi)$. Since the kernel of a linear map from a vector space $V$ to another vector space $W$ is a vector subspace, the claim follows (using the Rank-Nullity Theorem).

Now, I shall give you a good map $\varphi$. Define $$\varphi(p):=\big(p(+1)-p(-1),p(+2)-p(-2),p(+3)-p(-3),\ldots,p(+m)-p(-m)\big)\,,$$ where $m:=\left\lfloor\frac{n}{2}\right\rfloor$. It is easy to see that $\varphi$ is linear. It is also surjective since $$\varphi(q_j)=e_j\,,$$ where $e_1,e_2,\ldots,e_m\in\mathbb{R}^m$ are the usual standard basis vectors, and $$q_j(x):=\frac{\prod\limits_{i\in[m]\setminus\{j\}}\,\left(x^2-i^2\right)}{\prod\limits_{i\in[m]\setminus\{j\}}\,\left(j^2-i^2\right)}\in \mathbb{R}[x]\text{ for }j=1,2,\ldots,m\,.$$ Here, $[m]:=\{1,2,\ldots,m\}$.

The next task is to show that $S=\ker(\varphi)$. Clearly, by the definition of $S$, we see that $S\subseteq \ker(\varphi)$. We need to prove that $\ker(\varphi)\subseteq S$ as well. Let $p$ be an arbitrary element of $\ker(\varphi)$. Define $$f_p(x):=p(+x)-p(-x)\in\mathbb{R}[x]\,.$$ Thus, the roots of $f_p(x)$ are $0,\pm1,\pm2,\ldots,\pm m$. Consequently, $f_p$ has at least $$2m+1=2\left\lfloor\frac{n}{2}\right\rfloor+1\geq 2\left(\frac{n-1}{2}\right)+1=n$$ roots. As the degree of $f_p$ is at most the degree of $p$, which is $n-1$, we conclude that $f_p$ is identically zero. That is, $p(+x)=p(-x)$ identically, whence $p(+\alpha)=p(-\alpha)$ for all $\alpha\in\mathbb{R}$, so that $p\in S$.

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Alternatively, just observe that $S$ is spanned by the $\left\lceil\dfrac{m}{2}\right\rceil$ linearly independent polynomials $$1,x^2,x^4,\ldots,x^{2\,\left\lceil\frac{n}{2}\right\rceil-2}\,.$$ That is, every nonzero term in a polynomial $p\in S$ must have an even degree.

Answer 3

I claim that

$S = \{ p(x) \in P_n(x) \mid \forall \alpha \in \Bbb R, \; p(\alpha) = p(-\alpha) \} \tag 1$

is precisely the set

$E = \left \{ p(x) = \displaystyle \sum_0^{n - 1} p_i x_i \in P_n(x) \mid p_i = 0, i \; \text{odd} \right \}; \tag 2$

that is,

$S = E, \tag 3$

which means the polynomials in $P_n(x)$ which satisfy $p(\alpha) = p(-\alpha)$ are precisely those in which occur only even powers of $x$; for the present purposes we count the zero polynomial $\mathbf 0$ as having only even-degree terms, which makes sense insofar as $\mathbf 0$ is the constant polynomial whose value is $0$ for every $\alpha$:

$\mathbf 0(\alpha) = 0, \forall \alpha \in \Bbb R; \tag 4$

furthermore,

$\mathbf 0(\alpha) = 0 = \mathbf 0(-\alpha), \; \forall \alpha \in \Bbb R; \tag 5$

thus we see that

$\mathbf 0 \in S. \tag 6$

We may prove the assertion (3) as follows: first off, it is easy to see that

$E \subset S, \tag 7$

since even degree terms are of the form $ax^{2i}$, $a \in \Bbb R$; we have

$a\alpha^{2i} = a(\alpha^2)^i = a((-\alpha)^2)^i = a(-\alpha)^{2i}; \tag 8$

since any element of $E$ is the sum of such expressions, we see that (7) must bind.

Now suppose

$p(x) \in S, \tag 9$

and write

$p(x) = \eta(x) + \omega(x), \tag{10}$

where $\eta(x)$ consists of the even-degree monomials comprising $p(x)$, and $\omega(x)$ the odd; then for any $\alpha \in \Bbb R$ we have

$\eta(\alpha) + \omega(\alpha) = p(\alpha) = p(-\alpha) = \eta(-\alpha) + \omega(-\alpha) = \eta(\alpha) - \omega(\alpha), \tag{11}$

whence

$2 \omega(\alpha) = 0 \Longrightarrow \omega(\alpha) = 0; \tag{12}$

since $\omega(\alpha) = 0$ for every $\alpha \in \Bbb R$, we conclude that

$\omega(x) = 0, \tag{13}$

whence

$p(x) = \eta(x) \tag{14}$

has only terms of even degree; thus we see that

$S \subset E, \tag{15}$

so in fact

$S = E \tag{16}$

as claimed. Now $S = E$ is clearly a subspace of $P_n(x)$: the sum of two polynomials with only even-degree monomials obviously satisfies the same criterion, as does any product of such a polynomial with a scalar; and $\mathbf 0 \in E$ as we have seen.

And thats the $\alpha$ and $\omega$ of it!