Proving that $T_0, T_1, T_2, ...$ are basis of $\mathbb{R}[x]$

Question

I am given that the Chebyshev polynomial $T_n(x) \in \mathbb{Q}[x]$ is a polynomial such that $T_0(x) = 1$, $T_1(x) = x$ and for $n \ge 2$,

$T_n(x) = 2xT_{n-1}(x)-T_{n-2}(x)$

Now, I am supposed to show that $T_0, T_1, T_2, ...$ are a basis of real vector space $\mathbb{R}[x]$.

Basically, we want to show that $T_0, T_1, T_2, ...$ are linearly independent and span $\mathbb{R}[x]$. I have no clue how to start. I should use $deg(T_n) = n$ for span, but I am not sure how to show linear independence. Could someone help? Thanks.

Also, I want to prove this without any mentioning of inner product or orthogonality and orthonormality.

Answer 1

By induction, the degree of $T_n$ is $n$ for every non-negative integer number $n$. So, you can prove inductively that $T_0,\ldots,T_n$ spawn the vector subspace $\mathbb{R}_n[x]$ of polynomials of degree less or equal to $n$. We know that $\{1,x,\ldots,x^n\}$ is a basis of $\mathbb{R}_n[x]$ so by dimension you get the linear independence of $\{T_0,\ldots,T_n\}$.

Answer 2

Try induction on the degree part. Polynomials of different degree will be linearly independent for obvious reasons. Then spanning will follow from the degree argument.