Do $\{T_{n}\left(\frac{x^2}{2}-1\right)\}_{n=1}^\infty$ form a basis for the even degree polynomials? As we know, Chebyshev polynomials form a complete set of independent functions, i.e. they form a basis for the set of polynomials.
Let us consider a class of shifted Chebyshev polynomials of the first kind, e.g. $T_{n}\left(\frac{x^2}{2}-1\right)$. They generates independent polynomials, but they do not form a basis for the polynomials because, for example, all the $T_{n}\left(\frac{x^2}{2}-1\right)$ in this case are of even degree, so polynomials of odd degree cannot be formed from a linear combination of them.
My question is: do $\{T_{n}\left(\frac{x^2}{2}-1\right)\}_{n=1}^\infty$ form a basis for the even degree polynomials? Do $\{T_{1}\left(\frac{x^2}{2}-1\right),T_{2}\left(\frac{x^2}{2}-1\right),T_{4}\left(\frac{x^2}{2}-1\right),T_{8}\left(\frac{x^2}{2}-1\right),T_{16}\left(\frac{x^2}{2}-1\right),...,\}$ (and so on for even index) form a basis for the even degree polynomials?
 A: Yes, they form a basis of the space of even degree polynomial. Even more generally, if $(P_n)_{n \ge 0}$ is a basis of the space of polynomials, then $\left(P_n\left(\frac{X^2}{2}-1 \right)\right)_{n \ge 0}$ is a basis of the space of even degree polynomial :
To prove this, denote $E$ the space of odd degree polynomials, and observe that the linear map
$$\begin{matrix}
f : & \mathbb{R}[X] & \longrightarrow & E\\
& P(X) & \longmapsto & P \left(\frac{X^2}{2}-1 \right)\end{matrix}$$
is an isomorphism and hence it maps a basis of $\mathbb{R}[X]$ to a basis of $E$.
Indeed, let's prove this is indeed an isomorphism. First observe that $f$ does indeed map into $E$ because $P \left(\frac{X^2}{2}-1 \right)$ is always even for all $P$.
Injectivity : if $f(P) = 0$, then we have $\forall x \in \mathbb{R}, P \left(\frac{x^2}{2}-1 \right) = 0$, which implies that $P$ is zero on $[-1,\infty[$ so $P = 0$.
Surjectivity : let $Q(X) \in E$ be an even degree polynomial. We can write it as $Q(X) = \sum_{k = 0}^{s} a_{2k} X^{2k}$. So by setting $R(X) = \sum_{k = 0}^{s} a_{2k} X^{k} \in \mathbb{R}[X]$, we get $Q(X) = R\left(X^2\right)$. And thus by setting $P(X) = R(2(X+1)) \in \mathbb{R}[X]$, we get $R(X) = P\left(\frac{X}{2} - 1 \right)$ so
$$Q(X) = R\left(X^2\right) = P\left(\frac{X^2}{2} - 1 \right)$$
which proves that $Q = f(P)$.
