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

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.
