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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 ${\textbf q}_1, {\textbf q}_2, {\textbf q}_3$.

scatter plot of eigenvalues

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.

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$A:= \big(\begin{array}{c:c:c}2{\textbf q}_1 & -{\textbf q}_2 & -{\textbf q}_3\end{array} \big)$.

$ 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.

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    $\begingroup$ I know that the eigenvalues of $A$ are $4,1,1$. Does this imply my desired bound? Which property do I need for that? $\endgroup$ – GDumphart Jan 5 '16 at 14:27
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    $\begingroup$ Alright, I think I got it. With the given reasoning, $D^{-1} A^T$ is orthogonal too and thus preserves norm. $A$ and $A^T$ have the same eigenvalues. Let $\lambda$ be an eigenvalue of $A^T$ with unit eigenvector $v$, i.e. $A^T v = \lambda v$. Or rather $D^{-1} A^T v = \lambda D^{-1} v$. Taking the magnitude gives $1 = |\lambda| \cdot \|D^{-1} v\| \leq |\lambda| \cdot \|v\| = |\lambda|$, which is the bound. Please let me know if you have a more elegant approach. I'll check the stuff again in the evening and then accept your answer. Thank you very much so far. $\endgroup$ – GDumphart Jan 5 '16 at 15:40
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    $\begingroup$ I have to apologize, I did not think this through, and I now see where the problem is. I thought you could easily reason with the equation $AD^{-1}W = WE$ where $W$ is a matrix of eigenvectors (as columsn) and $E$ is a diagonal matrix of the corresponding eigenvalues. But it is not as simple as I thought, because I thought we had $EW$ on the RHS. But I'm happy that you could save the proof by using the norms. For me the original statement was intuitively clear because you still had rotation matrix, that just aditionaly stretches one dimension by a factor of $2$. (Greetings from Basel=) $\endgroup$ – flawr Jan 5 '16 at 19:16
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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 $|\lambda| \geq 1$.

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