Eigenvalue lower bound for this simple matrix For orthonormal vectors ${\textbf q}_1, {\textbf q}_2, {\textbf q}_3 \in \mathbb{R}^3$, I want to show that the matrix
$$\big(\begin{array}{c:c:c}2{\textbf q}_1 & -{\textbf q}_2 & -{\textbf q}_3\end{array} \big) \in \mathbb{R}^{3 \times 3}$$
has eigenvalues $\lambda_k$ fulfilling $1 \leq |\lambda_k| \leq 2$. Since the determinant is $2$, it suffices to show $1 \leq |\lambda_k|$.
I know this is true empirically: the following scatter plot shows $\lambda_1, \lambda_2, \lambda_3$ in the $\mathbb{C}$-plane for many random ${\textbf q}_1, {\textbf q}_2, {\textbf q}_3$.

How can I tackle this problem? I feel like this could have a very short answer, using a suitable property. I tried expressing ${\textbf q}_1, {\textbf q}_2, {\textbf q}_3$ with Euler angles and explicitly writing the characteristic polynomial in order to show its roots are on the discussed ring (with the hope of trigonometric simplifications), but to no avail. Thank you in advance.

$A:= \big(\begin{array}{c:c:c}2{\textbf q}_1 & -{\textbf q}_2 & -{\textbf q}_3\end{array} \big)$.
$ A^t A = \begin{pmatrix}4 & & \\ & 1 & \\ & & 1\end{pmatrix} =:D^2$ for $D=\begin{pmatrix}2 & & \\ & 1 & \\ & & 1\end{pmatrix}$.
$\implies I = D^{-1} A^t A D^{-1}= (AD^{-1})^tAD^{-1}$
This means that $AD^{-1}$ is orthogonal, and has therefore roots of unity as eigenvalues.
