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Problem 37: All polynomials in x over R with constant term zero

Problem 37: Is the following set form Vector space over reals?

All polynomials in x over R with constant term zero

Solution: Let $\Vfgpoly$,
where $f(0)=a_0=0$ and $g(0)=b_0=0$

I. Properties under addition
i. Closure Property
Let $f(x),g(x)$ is a real polynomial s.t. $f(0)=g(0)=0$.
Then $f(x)+g(x) = \polyab$
$f(0)+g(0) = a_0+b_0 = 0+0 =0 $
$\therefore f(x)+g(x) \in \V$
$\Rightarrow \V$ is closed under addition.

ii. Associativity:
Let $\Vfghpoly$
s.t $f(0)=a_0=0, g(0)=b_0=0,h(0)=c_0=0$
\begin{align*} \left[f(x)+g(x)\right]+h(x) = \left[\polya{a}+\polya{b}\right]+\polya{c}\\ =\polyab+\polya{c}\\ =\polyabc\\ =\polyabcc\\ \because (a_k+b_k)+c_k = a_k+(b_k+c_k), a_k, b_k, c_k \in \F\\ =\polya{a}+\polyax{(b_k+c_k}\\ =\polya{a}+\left[\polya{b}+\polya{c}\right]\\ =f(x)+\left[g(x)+h(x)\right]\\ \end{align*} $\Rightarrow$ Addition is associative.

iii. Existence of additive identity:
For each \begin{align*} \Vfpoly{f}{a} \in \V\\ O(x) = 0+0x+0x^2+...+0x^n+...\\ =\polyzero \in V\\ \end{align*} Now \begin{align*} f(x)+O(x)=\polya{a}+\polyzero\\ =\polyazero\\ =\polya{a} = f(x)\\ \end{align*} and \begin{align*} O(x)+f(x)=\polyzero+\polya{a}\\ =\polyazeroo\\ =\polya{a} = f(x)\\ \therefore f(x)+O(x)=f(x)=O(x)+f(x) \forall f(x) \in \V\\ \end{align*} $\Rightarrow O(x)$ is the additive identity.

iv. Existence of additive inverse:
For each \begin{align*} \Vfpoly{f}{a} \in \V\\ -f(x)=\polyax{(-a_k)} \in \V\\ \because a_k \in \F \Rightarrow -a_k \in \F\\ \end{align*} Now \begin{align*} f(x)+(-f(x))=\polya{a}+\polyax{(-a_k)}\\ =\polyax{\left[a_k+(-a_k)\right]}\\ =\polyzero = O(x)\\ \end{align*} and \begin{align*} (-f(x))+f(x)=\polyax{(-a_k)}+\polya{a}\\ =\polyax{\left[(-a_k)+a_k\right]}\\ =\polyzero = O(x)\\ \therefore f(x)+(-f(x))=O(x)=(-f(x))+f(x) \forall f(x) \in \V\\ \end{align*} $\Rightarrow -f(x)$ is the additive inverse.

v. Commutative Law:
Let \begin{align*} \Vfgpoly\\ \end{align*} Now \begin{align*} f(x)+g(x)=\polya{a}+\polya{b}\\ =\polyab\\ =\polyba\\ =\polya{b}+\polya{a}\\ = g(x)+f(x)\\ \end{align*} $\Rightarrow$ addition is commutative.

II. Properties under Scalar Multiplication
vi. Closure under Scalar Multiplication
Let $a\in \R$ and $\Vfpoly{f}{a} \in \V$ \begin{align*} af(0)=a(0)=0\\ \therefore af(x) \in \V \end{align*} $\Rightarrow \V$ is closed under Scalar Multiplication.

vii. Distributivity of Addition of Vectors
Let $\alpha\in \R$ and $f(x), g(x) \in \V$ \begin{align*} \alpha\left[f(x)+g(x)\right]= \alpha\left[\polya{a}+\polya{b}\right]\\ = \alpha\left[\polyab\right]\\ = \polyabalpha{\alpha}\\ = \polyabalphaa{\alpha}\\ =\polyaalpha{\alpha}{a}+\polyaalpha{\alpha}{b}\\ =\alpha(\polya{a})+\alpha(\polya{b})\\ =\alpha f(x)+\alpha g(x)\\ \end{align*} $\Rightarrow $ Distributivity of Addition of Vectors exists.

viii.Distributivity of Addition of Real Numbers
Let $\alpha, \beta\in \R$ and $f(x)\in \V$ \begin{align*} (\alpha+\beta)f(x)= (\alpha+\beta)\polya{a}\\ = \polyaalpha{(\alpha+\beta)}{a}\\ = \polyax{\left[\alpha a_k+\beta a_k\right]}\\ =\polyaalpha{\alpha}{a}+\polyaalpha{\beta}{a}\\ =\alpha(\polya{a})+\beta(\polya{a})\\ =\alpha f(x)+\beta f(x)\\ \end{align*} $\Rightarrow $ Distributivity of Addition of Real Numbers exists.

ix.Associativity of multiplication
Let $\alpha, \beta\in \R$ and $f(x)\in \V$ \begin{align*} (\alpha\beta)f(x)= (\alpha\beta)\polya{a}\\ = \polyax{((\alpha\beta)a_k)}\\ = \polyax{(\alpha(\beta {a_k}))}\\ = \alpha \left[\polyaalpha{\beta}{a}\right]\\ = \alpha \left[\beta\polya{a}\right]\\ =a(bf(x)) \end{align*} $\Rightarrow$ Associativity of multiplication exists.

x. Unity element multiplication
Let 1 be unity element of field $\F$ and $f(x)\in \V$
Now \begin{align*} 1.f(x)= 1.\polya{a}\\ =\polyax{(1.{a_k})}\\ =\polya{a}\\ = f(x) \end{align*} $\Rightarrow 1.f(x) = f(x) \forall f(x) \in \V$.

Hence, $\V$ is a vector space over $\R$.

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