Tridiagonal matrix with charcteristic polynomial is a polynomial $T_k(x)$ given recursively

Question

Let the polynomial sequence $T_k(x)$ as follow:

$T_0(x)=1$, $T_1(x)=x$ and $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$ for all $n\geq 1$.

Show the following:

1) For every $n\geq 2$ find a tridiagonal matrix $A_n$ so that its characteristic polynomial is $T_n(x)$.

2) Show that $A_n$ is similar to symmetric tridiagonal matrix.

The symmetric tridiagonal are the matrix in this link https://en.wikipedia.org/wiki/Tridiagonal_matrix

I showed one recursive formula for the polynomial characteristic of a tridiagonal matrix but i not can using this for to show the points 1 and 2 before.

I wait that can you help me with hints or any thing...

Answer 1

The $(T_n)$ are chebyshev polynomials, famous example of orthogonal polynomials. Their highest degree term is $2^nx^n$.

cf. https://en.wikipedia.org/wiki/Chebyshev_polynomials

Then $T_n$ cannot be a characteristic polynomial, but $U_n=2^{-n}T_n$ can.

The associated recurrence relation is : $U_{n+1}=xU_n-(1/4)U_{n-1}$.

A matrix $A_n=[a_{i,j}]$ which works is the following symmetric tridiagonal matrix:

$a_{i,i}=0,a_{i,i+1}=a_{i,i-1}=1/2$.