find power of orthonormal matrix to get identity matrix

Question

I have this orthonormal matrix:

$ Q =\frac{1}{9} \left(\begin{matrix} 4 & 8 & -1 \\ -4 & 1 & -8 \\ -7 & 4 & 4 \end{matrix}\right)$

If I calculate $Q^4$, I get the identity matrix. Is there a way to calculate the power $k$ of $Q^k$ to get the identity?

Answer 1

By computing $Q^TQ=I_3$ we see that $Q$ is orthogonal. Hence $Q$ is similar to

$$R_\theta=\begin{pmatrix}\pm1&0&0\\ 0&\cos\theta&-\sin\theta\\ 0&\sin\theta&\cos\theta \end{pmatrix}$$

Let $C_i$ the $i$-th column of $Q$. We have $C_1\wedge C_2=C_3$ so $Q$ is a rotation i.e. positive orthogonal matrix, hence $+1$ is an eigenvalue and from the trace of $Q$ we get $\cos\theta=0$ and so $\sin\theta=\pm1$. Now we see that $R_\theta^4=I_3$ hence $Q^4=I_3$.

Answer 2

I assume you want the least power $k$ so that $Q^k=I_3$. Given that $Q^4=I_3$, we know that $1\le k\le 4.$ On the other hand, if $k=3$, we would get that $I_3=Q^4=Q^3Q=I_3Q=Q$, which implies that $Q=I_3$, a contradiction. So, if $k\ne 4$, then $k=2$. So, just compute $k=2$. If $Q^2=I_3$, you are done. If $Q^2\ne I_3$, then the least $k$ is $4$.

Of course, another way to see this is that in any group $G$, if $g^\ell=e$, then $\operatorname{ord}(g)\mid \ell$. Here, the group is $GL_3(\mathbb{R})$.