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: $A:= \big(\begin{array}{c:c:c}2{\textbf q}_1 & -{\textbf q}_2 & -{\textbf q}_3\end{array} \big)$.
$ 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.
A: With the help of a colleague, I now got an answer which I think is reduced to the bare minimum. Still neither super short nor elegant, but the steps are basic and get the job done. First write
$$\big(\begin{array}{c:c:c}2{\textbf q}_1 & -{\textbf q}_2 & -{\textbf q}_3\end{array} \big) = {\textbf Q} \left(\begin{array}{ccc} 
2 && \\ & -1 & \\ && -1
\end{array} \right) \ .$$
The eigenvalue problem is
$${\textbf Q} \left(\begin{array}{ccc} 
2 && \\ & -1 & \\ && -1
\end{array} \right) {\textbf v} = \lambda {\textbf v} \ .$$
The magnitude of the right-hand side is $|\lambda| \cdot \|{\textbf v}\|$. The magnitude of the left-hand side is
$$\left\|{\textbf Q} \left(\begin{array}{ccc} 
2 && \\ & -1 & \\ && -1
\end{array} \right) {\textbf v} \right\| 
= \left\|\left(\begin{array}{ccc} 
2 && \\ & -1 & \\ && -1
\end{array} \right) {\textbf v} \right\| 
= \left\|\left(\begin{array}{r} 
2v_1 \\ -v_2 \\ -v_3
\end{array} \right) \right\| $$
$$= \sqrt{4v_1^2 + v_2^2 + v_3^2} \geq \sqrt{v_1^2 + v_2^2 + v_3^2} = \|{\textbf v}\| \ .$$
So $|\lambda| \cdot \|{\textbf v}\| \geq \|{\textbf v}\|$ gives the lower bound $|\lambda| \geq 1$.
