Series of characteristic polynomials Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $-1$, e.g.
$$
M_4= \begin{pmatrix}
               2 & -1 & 0 & 0 \\
               -1 & 2 & -1 & 0\\
               0 & -1 & 2 & -1\\
               0 & 0 & -1 & 2\\
              \end{pmatrix} 
$$
I've found out that the characteristic polynomials are
$$
\begin{cases}
P_1(x)=2-x\\
P_2(x)=(2-x)^2-1\\
P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x)
\end{cases}
$$
Or with a variable change
$$
\begin{cases}
Q_1(y)=y\\
Q_2(y)=y^2-1\\
Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y)
\end{cases}
$$
Looking at the first 8 $P_n$

I see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.


*

*How can I prove that all eigenvalues are between 0 and 4?

*Are these polynomials known (have a name)?

*How can I prove that the polynomial are sandwitched between
$$
\frac{1}{x}+\frac{1}{4-x}\quad\text{and}\quad
-\frac{1}{x}-\frac{1}{4-x}
$$


 A: From a very attenuated literature search, I see that the Lucas polynomials of the second kind obey the recursion
$$ L_{k+1}(x) = x \ L_k(x) - L_{k-1}(x), \quad  L_1(x) = 1 \ , L_2(x)=x$$
This matches the recursion for $Q_n,$ with an index shift of 1.  The one reference I've seen (I don't have access to journal papers from home) states that this polynomial set obeys the relation
$$L_k(2 \cos(\theta) ) = 2 \cos( k \ \theta) $$
This functional relationship would explain the location of the zeros, and likely the envelope property as well.  I suggest that the proposer start his research with 'Lucas polynomials of the second kind.'
A: The Chebyshev polynomials of the second kind satisfy the recurrence relation
$$
\begin{cases}
U_0(y) = 1 \\
U_1(y)=x\\
U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y)
\end{cases}
$$
so that $Q_n(y) = U_n(y/2)$ and $P_n(x) = U_n(1-x/2)$.
The zeros of $U_n$ are
$$
 y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n
$$
in the range $(-1, 1)$, so that
$$
 x_k = 2 - 2\cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n
$$
are the zeros of $P_n$ in the range $(0, 4)$.
Also for $|x| < 1$
$$
U_n(x) = \frac{\sin((n+1)\arccos(x))}{\sqrt{1-x^2}} 
$$
which implies
$$
|U_n(x)| \le \frac{1}{\sqrt{1-x^2}} 
$$
and therefore
$$
| P_n(x)| \le \frac{2}{\sqrt{x(4-x)}} \le \frac 1x + \frac{1}{4-x}
$$
for $0 < x < 4$, the last estimate follows from the inequality between harmonic and geometric mean.
