How to solve $K^n = I$ I'm supposed to show that there exists an n such that given
\begin{equation*}
K = 
\begin{pmatrix}
0 & 0 & i \\
-i & 0 & 0 \\
0 & -1 & 0
\end{pmatrix}
\end{equation*}
$K^n = I$ where $I$ is the identity.
So
\begin{equation*}
I = 
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\end{equation*}
It turns out n = 6 but how does one arrive at this efficiently? What I mean is that I don't know how to immediately solve $K^n = I$
 A: You can "see" without writing anything down that $\vec{e_1}\mapsto -i\vec{e_2}\mapsto i\vec{e_3}\mapsto -\vec{e_1}\mapsto i\vec{e_2}\mapsto -i\vec{e_3}\mapsto \vec{e_1}$.
So $K^6\vec{e_1}=\vec{e_1}$.
Scale that chain by the appropriate power of $i$ to see that similar cycles hold with $\vec{e_2}$ and $\vec{e_3}$.
A: If you were to take the direct approach and compute successive powers of $K$, you would find that $K^3 = -I$. Thus, for $n > 3$, we have $K^n = -K^{n-3}$. Because $K$ and $K^2$ are not the identity matrix, we can therefore conclude that the smallest $n$ for which $K^n = I$ is $n=6$, since
$$
K^6 = - K^{3} = -(-I) = I.
$$
An alternative is to note that $K$ looks "almost like" a companion matrix. Note that $K$ is similar to the matrix
$$
M = \pmatrix{i\\&1\\&&1} \ K\ 
\pmatrix{i\\&1\\&&1}^{-1} = -\pmatrix{0&0&1\\1&0&0\\0&1&0}.
$$
Note that $-M$ is the companion matrix associated with the polynomial $x^3 - 1$. Conclude that $M^3 = -I$, which implies that $K^3 = -I$, as before. From there, proceed in the same way.
