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Problem 53: Prove that $P_n$ is a vector space

Problem 53: Prove that $P_n$ is a vector space<\p>

Solution:
The Axioms of a Vector Space
The following properties must hold for all $ u,v,w \in \V$ and $a,b \in \R$:
Closure Properties
$(A1) u+v \in \V$.
$(A2) av \in \V$.
Properties of Addition
$(A3) u+v=v+u$.
$(A4) u+(v+w)=(u+v)+w$.
$(A5)$ There is an element $\mathbf{0} \in \V$ such that $\mathbf{0}+v=v$ for all $v \in \V$.
$(A6)$ Given an element $v \in \V$, there is an element $−v \in \V$ such that $v+(−v)=\mathbf{0}$.
Properties of Scalar Multiplication
$(A7) a(bv)=(ab)v$.
$(A8) a(u+v)=au+av$.
$(A9) (a+b)v=av+bv$.
$(A10) 1v=v$ for all $v \in \V$.

Prove A1:
If $a_0 + a_1x + · · · + a_nx^n$ and $b_0 + b_1x + · · · + b_nx^n$ belong to $P_n$, then $(a_0 + a_1x + · · · + a_nx^n) + (b_0 + b_1x + · · · + b_nx^n)$
$= (a_0 + b_0) + (a_1 + b_1)x + · · · + (a_n + b_n)x^n$, which again belongs to $P_n$. Therefore, $P_n$ is closed under addition.

Prove A2:
If $a_0 + a_1x + · · · + a_nx^n$ and $r$ is a scalar, then
$r\cdot(a_0 + a_1x + · · · + a_nx^n) = (ra_0) + (ra_1)x + · · · + (ra_n)x^n$,
which again belongs to $P_n$. Therefore, $P_n$ is closed under scalar multiplication.

Prove A3:
Let $p(x) = a_0 + a_1x + · · · + a_nx^n$ and $q(x) = b_0 + b_1x + · · · + b_nx^n$ belong to $P_n$.
Then
$p(x) + q(x) = (a_0 + a_1x + · · · + a_nx^n) + (b_0 + b_1x + · · · + b_nx^n)$
$= (a_0 + b_0) + (a_1 + b_1)x + · · · + (a_n + b_n)x^n$
$= (b_0 + a_0) + (b_1 + a_1)x + · · · + (b_n + a_n)x^n$
$= (b_0 + b_1x + · · · + b_nx^n) + (a_0 + a_1x + · · · + a_nx^n)$
$= q(x) + p(x),$
so $P_n$ satisfies commutativity under addition.

Prove A4:
Let $p(x) = a_0 +a_1x+· · ·+a_nx^n, q(x) = b_0 +b_1x+· · ·+b_nx^n,$ and $r(x) = c0 +c1x+· · ·+cnx^n$ belong to $P_n$. Then
$\left[p(x) + q(x)\right] + r(x) = \left[(a_0 + a_1x + · · · + a_nx^n) + (b_0 + b_1x + · · · + b_nx^n)\right] + (c_0 + c_1x + · · · + c_nx^n)$
$= \left[(a_0 + b_0) + (a_1 + b_1)x + · · · + (a_n + b_n)x^n\right] + (c_0 + c_1x + · · · + c_nx^n)$
$= \left[(a_0 + b_0) + c_0\right] + \left[(a_1 + b_1) + c_1\right]x + · · · + \left[(a_n + b_n) + c_n\right]x^n$
$= \left[a_0 + (b_0 + c_0)\right] + \left[a_1 + (b_1 + c_1)\right]x + · · · + \left[a_n + (b_n + c_n)\right]x^n$
$= (a_0 + a_1x + · · · + a_nx^n) + \left[(b_0 + c_0) + (b_1 + c_1)x + · · · + (b_n + c_n)x^n\right]$
$= (a_0 + a_1x + · · · + a_nx^n) + \left[(b_0 + b_1x + · · · + b_nx^n) + (c_0 + c_1x + · · · + c_nx^n)\right]$
$= p(x) + \left[q(x) + r(x)\right],$
so $P_n$ satisfies associativity under addition.

Prove A5:
The zero vector is the zero polynomial $z(x) = 0 + 0x + · · · + 0x^n$, and it is readily verified that this
polynomial satisfies $z(x) + p(x) = p(x) = p(x) + z(x)$ for all $p(x) \in P_n$.

Prove A6:
The additive inverse of $p(x) = a_0 + a_1x + · · · + a_nx^n$ is
$−p(x) = (−a_0) + (−a_1)x + · · · + (−a_n)x^n$.
It is readily verified that $p(x) + (−p(x)) = z(x)$, where $z(x)$ is defined in A5.

Prove A7:
Let $r, s \in \R,$ and $p(x) = a_0 + a_1x + · · · + a_nx^n \in P_n$.
Then
$(rs)\cdot p(x) = (rs)\cdot(a_0 + a_1x + · · · + a_nx^n)$
$= \left[(rs)a_0\right] + \left[(rs)a_1\right]x + · · · + \left[(rs)a_n\right]x^n$
$= r\left[(sa_0) + (sa_1)x + · · · + (sa_n)x^n\right]$
$= r\left[s(a_0 + a_1x + · · · + a_nx^n)\right]$
$= r\cdot(s\cdotp(x))$,
which verifies the associativity of scalar multiplication.

Prove A8:
Let $r \in \R$, Let $p(x) = a_0 + a_1x + · · · + a_nx^n \in P_n$, and $let q(x) = b_0 + b_1x + · · · + b_nx^n \in P_n$.
Then
$r \cdot(p(x) + q(x)) = r \cdot ((a_0 + a_1x + · · · + a_nx^n) + (b_0 + b_1x + · · · + b_nx^n))$
$= r \cdot \left[(a_0 + b_0) + (a_1 + b_1)x + · · · + (a_n + b_n)x^n\right]$
$= \left[r(a_0 + b_0)\right] + \left[r(a_1 + b_1)\right]x + · · · + \left[r(a_n + b_n)\right]x^n$
$= \left[(ra_0) + (ra_1)x + · · · + (ra_n)x^n\right] + \left[(rb_0) + (rb_1)x + · · · + (rb_n)x^n\right]$
$= \left[r(a_0 + a_1x + · · · + a_nx^n)\right] + \left[r(b_0 + b_1x + · · · + b_nx^n)\right]$
$= r \cdot p(x) + r \cdot q(x)$,
which verifies the distributivity of scalar multiplication over vector addition.

Prove A9:
Let $r, s\in \R$ and $p(x) = a_0 + a_1x + · · · + a_nx^n \in P_n$.
Then
$(r + s) \cdot p(x) = (r + s) \cdot (a_0 + a_1x + · · · + a_nx^n)$
$= \left[(r + s)a_0\right] + \left[(r + s)a_1\right]x + · · · + \left[(r + s)a_n\right]x^n$
$= \left[ra_0 + ra_1x + · · · + ra_nx^n\right] + \left[sa_0 + sa_1x + · · · + sa_nx^n\right]$
$= r(a_0 + a_1x + · · · + a_nx^n) + s(a_0 + a_1x + · · · + a_nx^n)$
$= r \cdot p(x) + s \cdot p(x)$,
which verifies the distributivity of scalar multiplication over scalar addition.

Prove A10:
We have
$1\cdot(a_0 + a_1x + · · · + a_nx^n) = a_0 + a_1x + · · · + a_nx^n$,
which demonstrates the unit property in $P_n$.

$\therefore$ The above verification of axioms A1-A10 shows that $P_n$ is a vector space.

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