Is $\beta$ a basis of vector space $V$

linear algebrasolution-verificationvector-spaces

Let $V=\{p\in\mathbb{R}[X]:\deg(p)\leq n\}$, knowing that $\{1,X,\dots,X^n\}$ is a basis of $V$, determine whether $\beta=\{1,X,X^2+1,X^3+X,\dots,X^n+X^{n-2}\}$ is a basis of $V$.
Consider: $\quad c_0+c_1X+c_2(X^2+1)+\dots+c_n(X^n+X^{n-2})=0$
$\implies (c_0+c_2)+(c_1+c_3)X+(c_2+c_4)X^2+\dots+(c_n+c_{n-2})X^{n-2}+c_{n-1}X^{n-1}+c_nX^n=0$
Because of the definition of the zero polynomial, it must follow, that all coefficients $\quad c_0,c_1,\dots,c_n=0$, meaning $\beta$ is linearly independent.
It is given that $\{1,X,\dots,X^n\}$ is a basis of $V$, so $\dim(V)=\dim(\{1,X,\dots,X^n\})=n+1$. Because $|\beta|=|\{1,X,\dots,X^n\}|,\;\dim(V)=\dim(\beta)$. Therefore $\beta$ is a basis of $V$.
Is this proof correct? Thank you for the help

Best Answer

Your proof is incomplete. After getting that$$(c_0+c_2)+(c_1+c_3)X+\cdots+(c_{n-2}+c_n)X^{n-2}+c_{n-1}X^{n-1}+c_nX^n=0,$$you can't jump right away to $c_0=c_1=\cdots=c_n=0$. Note that$$\left\{\begin{array}{l}c_0+c_2=0\\c_1+c_3=0\\\vdots\\c_{n-2}+c_n=0\\c_{n-1}=0\\c_n=0.\end{array}\right.$$Now,

  • From $c_n=0$ and $c_{n-2}+c_n=0$, you get that $c_{n-2}=0$.
  • From $c_{n-1}=0$ and $c_{n-3}+c_{n-1}=0$, you get that $c_{n-3}=0$.
  • $\vdots$
  • And so on, until you get that $c_0=0$.
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