Main Question
Say that $K$ and $S$ are two commuting matrices with a full set of real eigenvectors, where the dimension of the eigenspaces of the matrices may be greater than one.
$$KS=SK$$
It is a common theorem in linear algebra that $S$ and $K$ will share eigenvectors. The top answer in this question explains it beautifully:
Matrices commute if and only if they share a common basis of eigenvectors?
However, say that the eigenvectors of $K$ are real, while the eigenvectors of $S$ are complex.
In this case, will the matrices still share eigenvectors?
Or more specifically, will we allways be able to linearly combine the complex eigenvectors of $S$ to get the real eigenvectors of $K$?
Reason for Asking:
(Ignore the physics part if this doesn't ring a bell).
In physics textbooks deriving the normal-modes of coupled oscillators, its common practice to introduce a "symmetry" matrix transformation. Instead of finding the eigenvectors of the matrix describing the coupling of the system $K$, the eigenvectors of the symmetry matrix $S$ are found, and those correspond to the normal-modes of the actual system.
The theorem that's used to justify the fact that $K$ and $S$ share eigenvectors is that they always commute.
However, often the eigenvectors of $S$ are complex, while the eigenvectors of $K$ are real. In this case, the complex eigenvectors of $S$ are linearly combined to give real vectors, and those turn out to be eigenvectors of $K$.
Can someone explain/prove why this is? Or point me to some useful theorems about commuting matrices when the eigenvectors of one are complex, while the eigenvectors of the other are real?
Thank you!
Example:
Consider the two matrices:
$$K=\left(\begin{array}{ccc}
-2 & 1 & 1 \\
1 & -2 & 1 \\
1 & 1 & -2
\end{array}\right) S=\left(\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right)$$
$$$$
These two matrices commute, to give:
$$SK=KS=\left(\begin{array}{lll}
1 & -2 & 1 \\
1 & 1 & -2 \\
-2 & 1 & 1
\end{array}\right)$$
The eigenvectors of $S$ would be complex. We can solve for them by first solving for the eigenvalues, $\lambda_k$, of which there should be three.
$$\begin{array}{c}
\operatorname{det}\left|\mathrm{S}-\mathrm{\Lambda}_{\mathrm{k}}\right|=0 \\
\operatorname{det}\left[\begin{array}{ccc}
-\lambda_{k} & 1 & 0 \\
0 & -\lambda_{k} & 1 \\
1 & 0 & -\lambda_{k}
\end{array}\right]=0 \\
-\lambda_{k}^{3}=-1 \\
\lambda_{k}^{3}=1
\end{array}$$
And we need to solve for the cubic roots of $1 .$ There are three of those:
$$
\lambda_{0}=1, \lambda_{1}=\frac{2 \pi}{3}, \lambda_{2}=\frac{4 \pi}{3}
$$
If we assume that the first component of each of the eigenvectors is $1$, that gives us three distinct eigenvectors:
$$\lambda_{0}=1 \Rightarrow \overrightarrow{A^{(0)}}=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right], \lambda_{1}=e^{i \frac{2 \pi}{3}} \Rightarrow \overrightarrow{A^{(1)}}=\left[\begin{array}{c}1 \\ e^{i \frac{2 \pi}{3}} \\ e^{i \frac{4 \pi}{3}}\end{array}\right], \lambda_{2}=e^{i \frac{4 \pi}{3}} \Rightarrow \overrightarrow{A^{(2)}}=\left[\begin{array}{c}1 \\ e^{i \frac{4 \pi}{3}} \\ e^{i \frac{8 \pi}{3}}\end{array}\right]$$
If we write out the complex and real parts of $S$, we get:
$$\overrightarrow{A^{(0)}}=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]$$
$$\overrightarrow{A^{(1)}}=\left[\begin{array}{c}
1 \\
-0.5 \\
-0.5
\end{array}\right] + \left[\begin{array}{c}
0 \\
.866i\\
-.866i
\end{array}\right]$$
$$\overrightarrow{A^{(2)}}=\left[\begin{array}{c}
1 \\
-0.5 \\
-0.5
\end{array}\right] + \left[\begin{array}{c}
0 \\
-.866i\\
+.866i
\end{array}\right]$$
And, I'll leave it up to you to verify that the complex and real parts of each of those eigenvectors are eigenvectors of $K$ as well.
Best Answer
Your problem is virtually trivial when your write it properly. S is the celebrated cyclic permutation matrix, $$S=\left(\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right)~~\leadsto S^T=S^{-1},$$ with the famed eigenvalues and eigenvectors $$ 1, ~~\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]; ~~~~~ \omega, ~~\left[\begin{array}{l} 1 \\ \omega \\ \omega^2 \end{array}\right]; ~~~~~ \omega^2 , ~~\left[\begin{array}{l} 1 \\ \omega^2 \\ \omega \end{array}\right], $$ where $\omega=\exp (i2\pi/3)$, so $\omega^3=1$, and $1+\omega+ \omega^2=0$.
Now, $$ K= -2 {\mathbb I} +S + S^2, $$ so of course it manifestly commutes with S, and shares its 3 eigenvectors, now with eigenvalues 0, -3, -3, respectively.
The first is real, and you may take real and imaginary parts by considering the sum and difference of the latter two, and chuck the i of the imaginary piece. For this to work you need degeneracy, as you have here.