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

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

• Seems ok. Alternatively you can note that the matrix with 1's in the diagonal and 1's placed diagonally two places above the main diagonal is non-singular (because it's triangular with non-zero diagonal entries). Commented Aug 14, 2020 at 18:34
• Your proof is fine, I would just add a detail. Instead of saying “definition of the zero polynomial” I would say “as $1,X,...,X^n$ is a basis of V then $c_{0}+c_{2}=...=c_{n}=0$“.
– user723846
Commented Aug 14, 2020 at 18:35
• @JCAA I figured that after rearranging the coefficient, it has the form $p(x)=a_nx^n+...a_1x+a_0=0$ meaning it is a zero polynomial, and because of the definition of the zero polynomial, all coefficients must be zero. Commented Aug 14, 2020 at 18:35
• @Gio so because of the basis that is given, does it follow directly, that all $c_1, ..., c_n=0$ or is there a step in between? Commented Aug 14, 2020 at 18:37
• As $c_{n}=0$ and $c_{n}+c_{n-2}=0$ then $c_{n-2}=0$. You can do the same with $c_{n-1}=0$ and $c_{n-1}+c_{n-3}=0$ until you get that all $c_{i}=0$ for $i=0,...,n$.
– user723846
Commented Aug 14, 2020 at 18:44

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 $$c_0=0$$.
More simply, you may consider the determinant of $$\beta$$ in the standard basis:
$$\det\beta=\begin{vmatrix} 1 & 0 & 1 & 0 & \dots\dots & 0 \\ 0 & 1 & 0 & 1 & \dots\dots & 0 \\ 0 & 0 & 1 & 0 & \dots\dots & 0 \\ 0 & 0 & 0 & 1 & \dots\dots & 0 \\ \vdots & & & & \ddots& \vdots \\ 0 & 0 & 0 & 0 & \dots\dots & 1 \end{vmatrix}$$ It is an upper triangular determinant, so its value is the product of the diagonal element, $$1$$, which proves $$\beta$$ is a basis.