Without Calculation, why is $Q^{-1}$ a Matrix of Rotation

Question

Let $\ M=\frac{1}{9} \begin{pmatrix} -8 & 4 & 1 \\ -1 & -4 & 8 \\ 4 & 7 & 4 \end{pmatrix}$. Explain without calculation why $Q$ is invertible and $Q^{-1}$ is the matrix of a rotation. If this rotation has axis $\vec{w}$ and angle $\beta$, write down $\vec{w}$ and then explain how $\beta$ is related to $\alpha$, the angle of rotation of $Q$.

Progress:

So far through this question, I have shown that $Q$ is indeed a rotation matrix, as $-1$ is not an eigenvalue of $Q$, whereas $1$ is an eigenvalue of $Q$. Hence as $Q$ is a rotation matrix, det$(Q)=1\Rightarrow Q$ is invertible.

However, I don't understand why $Q^{-1}$ is also a rotation matrix. I thought that if $Q=Q^T$, where $T$ denotes the transpose of $Q$,then this could be one reason as $Q^T=Q^{-1}$ as $Q$ is orthogonal. But this is not the case. I am unsure of how to proceed.

Edit: Matrix has been corrected as noted by José Carlos Santos.

Answer 1

Note simply that if $Q$ is a rotation matrix of $\theta$ around a certain axis vector then also the inverse transformation expressed by $Q^{-1}$ is a rotation matrix of $-\theta$ around the same axis vector.

Answer 2

I don't know how to do this “without calculation”, but $Q^T.Q=\operatorname{Id}$ and therefore $Q$ is an orthogonal matrix. Besides, its only real eigenvalue is $1$ (with multiplicity $1$) and so $Q$ is indeed a rotation. The vector $(1,3,5)$ is an eigenvector and therefore $Q$ is the matrix of a rotation around the line defined by the origin and $(1,3,5)$.

Answer 3

It is sufficient to show det$(Q) = 1$ in order for $Q$ to be the matrix of a pure rotation. Some intuition for this is that it can be shown that for a $3 \times 3$ orthogonal matrix $X$, if det$(X) = 1$, then $X$ is similar to $$\left(\begin{matrix} 1 & 0 & 0\\ 0 & \text{cos}\theta & -\text{sin}\theta \\ 0 & \text{sin}\theta & \text{cos}\theta \end{matrix}\right)$$ where $\theta \in \mathbb{R}$. I.e there exists a (turns out, orthonormal, which is what you want) basis of $\mathbb{R}^3$ such that $X$ acts as a rotation of the coordinate vectors in such a basis about the $x$-axis through an angle $\theta$ (clockwise when looking at the $y$-$z$ plane from the positive x-axis).

Or, you can think of the determinant as how a linear transformation changes volumes, 3blue1brown has an excellent explanation of this, see https://www.youtube.com/watch?v=Ip3X9LOh2dk. Orthogonal matrices preserve volumes, though a determinant of $-1$ means space gets "inverted", i.e a rotation and a reflection. No squishing allowed!

Furthermore, det$(Q) \neq 0$, therefore $Q$ is invertible. Now, det$(Q^{-1}) = \text{det}(Q^T) = \text{det}(Q) = 1$, since $Q$ is orthogonal, and so $Q^{-1}$ is also the matrix of a pure rotation.

Geometrically, the transformation $Q^{-1} Q = I$ represents a rotation composed with another rotation, such that the net effect is nothing. This means that if we apply $Q^{-1} Q$ to a vector $\textbf{v} \in \mathbb{R}^3$, it gets rotated around the axis of rotation defined by $Q$ (which you can find by finding the eigenvectors corresponding to the eigenvalue $1$, as José already pointed out) by an angle $\alpha$, and then another rotation happens such that it ends up where it started.

What could that be? A rotation about the same axis as that of $Q$, by an angle $\beta = 2 n \pi - \alpha$, $n \in \mathbb{Z}$, or just $-\alpha$.