Let gamma be the real root of a polynomial equation of degree 9 with integer coefficients.

Question

Let γ be the real root of a polynomial equation of degree 9 with integer coefficients. Construct the matrix

$\left A = \right \left (\begin{array}{rrr} 2 & \gamma & 0 \\ \gamma & 2 & \gamma \\ 0 & \gamma & 2 \end{array} \right)$

With this information, is it possible to
i) derive all the possible values of γ so that A has all non-zero eigenvalues?
ii) calculate the necessary condition on γ so that all the eigenvalues of A are positive?

Solution

$\left A = \right \left (\begin{array}{rrr} 2 & \gamma & 0 \\ \gamma & 2 & \gamma \\ 0 & \gamma & 2 \end{array} \right)$

This means that A is a real, symmetric matrix.

i) Let’s find the eigenvalues of $|A - \lambda I| = 0$

$\left |A - \lambda I| = \right \left| \begin{array}{rrr} 2-\lambda & \gamma & 0 \\ \gamma & 2-\lambda & \gamma \\ 0 & \gamma & 2-\lambda \end{array} \right|$

$\begin{align*} \left = (2-\lambda).[(2-\lambda)^2 - \gamma^2] - \gamma.[\gamma.(2-\lambda)] \end{align*}$

$\begin{align*} \left = (2-\lambda)^3 - \gamma^2(2-\lambda) - \gamma^2(2-\lambda) = 0 \end{align*}$

$\begin{align*} \left = (2-\lambda)^2 - 2\gamma^2 = 0 \end{align*}$

$\begin{align*} \left (2-\lambda)^2 = 2\gamma^2 = 0 \end{align*}$

$\begin{align*} \left (2-\lambda) = 2\gamma^2 = 0 \end{align*}$

(1) $\begin{equation*} \left \lambda = 2 - \sqrt2\gamma \end{equation*}$

For (1) to yield positive values:

$\begin{align*} \left 2 - \sqrt2\gamma > 0 \end{align*}$

$\begin{align*} \left 2 > \sqrt2\gamma \end{align*}$

$\begin{align*} \left \gamma < \sqrt2 \end{align*}$

Thus, necessary condition is $\gamma < \sqrt2$ when γ is positive. But if $\gamma$ is negative, then it can be any value $\gamma \in \mathbb{R}$ .

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