Prove that these sets of polynomials have real and distinct roots. Can anyone tell me if the following set of polynomials have a special name?
$$P_{0}(x)=1,P_{1}(x)=x$$
$$P_{n}(x)=xP_{n-1}-P_{n-2}$$
The above gives:
$$P_{2}(x)=x^2-1;P_{3}(x)=x^3-2x;P_{4}(x)=x^4-3x^2+1;P_{5}(x)=x^5-4x^3+3x$$
So $P_{n}(x)$ has parity $(-1)^n$. I was trying to find out whether they are orthogonal, but couldn't find a suitable weight function. My main concern is to prove that $P_{n}(x)$ has n distinct real roots, all larger than or equal to -2.
 A: Your polynomials are indeed a special case of classical orthogonal polynomials. 
According to Abramowitz/Stegun 22.7.6 you have 
$$P_n(x) = S_n(x)= U_ n\left(\frac{x}{2}\right)$$
where $U_n(x)$ is the well-known Chebyshev polynomial of the second kind.
The weight function for the interval $(-2,2)$ is 
$$w(x)=\left(1-\frac{x^2}{4}\right)^{1/2}$$
And of course this means that the root are simple, distinct and located in the interval $(-2,2).$ For a proof  see e.g. my answer
Proof the Legendre polynomial $P_n$ has $n$ distinct real zeros .

The orthogonality of $P_n$ follows from the correspending property of $U_n$ and $\sin$ , see e.g. https://www.sciencedirect.com/science/article/pii/0377042793901485:
With $x=\cos\theta$ and $\sin \theta = (1-x^2)^{1/2}$ you have
$$U_n(\cos \theta) = \frac{\sin\big((n{+}1)\theta\big)}{\sin\theta}$$ 
so $$(1-x^2)^{1/2}U_n(x) = \sin\big((n{+}1)\theta\big)$$
(Although I did not see any fully formulated proof yet, maybe a direct proof from the recursion can be modelled after https://planetmath.org/orthogonalityofchebyshevpolynomialsfromrecursion)
A: In what follows the explicit expression for the roots of the polynomials $P_n(x)$ will be derived. The statement "the roots are all distinct, real and less than 2 by absolute value" follows immideately. 

Consider a family of $n\times n$ bidiagonal matrices:
$$\begin{align}
A^{(n)}_{ij}=&\delta_{i-j,1}+\delta_{j-i,1},
\end{align}\tag{1}$$
given below for $n=5$ as example:
$$
A^{(5)}=\begin{pmatrix}
0&1&0&0&0\\
1&0&1&0&0\\
0&1&0&1&0\\
0&0&1&0&1\\
0&0&0&1&0
\end{pmatrix}.
$$
Lemma 1. The eigenvalues of the matrix (1) are:
$$
\begin{align} \lambda_m=2\cos\frac{\pi m}{n+1},&
\text{with associated eigenvectors } u_{mk}=\sin\frac{\pi m}{n+1}k,
\end{align}\tag{2}
$$
where $m$ and $k$ run from 1 to $n$.
Though it would suffice for the proof to let the matrix $A$ act on the given vectors, we present below an extended "constructive" version. 
Assume the elements of an eigenvector $u$ have the form:
$$
u_k=e^{\alpha k}+ae^{-\alpha k},\tag{3}
$$
with some parameters $a$ and $\alpha$, which are to be found.
Obviously for all $k=2\dots(n-1)$
$$
(Au)_k=\left(e^{\alpha k}+ae^{-\alpha k}\right)\left(e^\alpha+e^{-\alpha}\right)
=\left(e^\alpha+e^{-\alpha}\right)u_k.\tag{4}$$
Thus it remains only to find such $a$ and $\alpha$ that the equation (4) is satisfied for $k=1$ and $k=n$ as well. 
For $k=1$:
$$
e^{\alpha 2}+ae^{-\alpha 2}=\left(e^\alpha+e^{-\alpha}\right)\left(e^\alpha+a e^{-\alpha}\right)
\Leftrightarrow
1+a=0.\tag{5}
$$
For $k=n$:
$$
e^{\alpha (n-1)}+ae^{-\alpha(n-1)}=\left(e^\alpha+e^{-\alpha}\right)\left(e^{\alpha n}+a e^{-\alpha n}\right)
\Leftrightarrow e^{\alpha (n+1)}+ae^{-\alpha(n+1)}=0.\tag{6}
$$
It follows: $a=-1$, $\alpha=\frac{\pi m}{n+1}i$, where $m$ is an integer number. Plugging the values into (3) and (4) one obtains (2).
As all $n$ eigenvalues are distinct, Lemma 1 is proved.
Lemma 2. The characteristic polynomials of negated matrix (1):
$$
Q_n(x)\equiv\left|A^{(n)}+x I^{(n)}\right|,
$$
where $I^{(n)}$ is $n\times n$ dimensional identity matrix, are the polynomials in question:
$$Q_n(x)=P_n(x)\tag{7}.$$
For $n=1$ and $n=2$ the equality (7) is obvious. Assume that (7) is valid for all $n<N$. Then it is valid for $n=N$ as well. 
Indeed, applying the Laplace expansion to matrix $A^{(N)}$ (with $N>2$) one readily obtains:
$$
Q_N(x)=x Q_{N-1}(x)-Q_{N-2}(x)\stackrel{I.H.}{=}x P_{N-1}(x)-P_{N-2}(x)=P_N(x).\tag{8}
$$
Thus, by induction Lemma 2 is proved.
Now, as the eigenvalues of a matrix are  exactly the roots of its characteristic polynomial,
Lemma 3. The roots of $P_n(\lambda)$ are:
$$
\lambda^{(n)}_m=2\cos\frac{\pi m}{n+1}, \quad m=1\dots n
$$
is a simple Corollary of Lemmas 1 and 2.
