How to determine the existence of certain common eigenvectors in a simpler way? I met such a linear algebra exercise, and I'm thinking about whether there's any trick to avoid brute-force-computation:
Let $V$ be a finite dimentional vector space over $\mathbb{C}$. Let $W:=V \otimes V \otimes V$, and
$T_{1,2}: W\rightarrow W$ are two linear mappings defined as:
$$T_1(x\otimes y \otimes z)=y\otimes x\otimes z,~~ T_2(x\otimes y \otimes z)=x\otimes z\otimes y.$$
Exercise: determine the pairs $(a_1,a_2)$ of eigenvalues of $T_1$ and $T_2$: $a_j$ is an eigenvalue of $T_j$, such that they share a common eigenvector $w\in W$ nonzero with $T_j(w_j)=a_j w_j$.
So I found that both of them can be diagonalised, with eigenvalues $1$ and $-1$: because we can find a basis for $T_1$ as follows
$$
\left\{\begin{array}{l}
\left(e_{i} \otimes e_{i}\right) \otimes e_{k} \\
\left(e_{i} \otimes e_{j}+e_{j} \otimes e_{i}\right) \otimes e_{k} \\
\left(e_{i} \otimes e_{j}-e_{j} \otimes e_{i}\right) \otimes e_{k}
\end{array}\right., (1\neq i<j\leq d:=dim(V))
$$
and similarly for $T_2$:
$$
\left\{\begin{array}{l}
\left( e_{k} \otimes (e_{i} \otimes e_{i}\right) \\
\left (e_{k} \otimes (e_{i} \otimes e_{j}+e_{j} \otimes e_{i}\right) \\
\left (e_{k} \otimes (e_{i} \otimes e_{j}-e_{j} \otimes e_{i}\right) 
\end{array}\right., (1\neq i<j\leq d:=dim(V))
$$
Now the question is: to determine when my choice has a common eigenvector. So obviously $(1,1)$ is correct. But is there any trick to "See" the cases $(1,-1),(-1,1),(-1,-1)$? I'm trying to find a method without too much "brute-force".
Any help/hint is welcome, thanks in advance! (And I guess there's some background behind this exercise. If you have any idea, please give me some clue, thanks!)
 A: Partial answer.
(I) The cases $(1,-1)$ and $(-1,1)$ don't occur.
To see this note that if $w$ is a $\lambda_i$-eigenvector of $T_i$ for $i=1,2$ then $w$ is a $\lambda_1 \lambda_2$-eigenvector of $S=T_1 T_2$. But $S:x\otimes y\otimes z\mapsto z\otimes x\otimes y$ has order $3$ so its eigenvalues lie in $\{1,\omega,\omega^2\}$ (where $\omega$ is a primitive cube root of unity).
(II) Provided $\dim V>2$ the case $(-1,-1)$ does occur.
To see this note that any common eigenvector must be a $1$-eigenvector of $S$. That suggests where we should look for these common eigenvectors.
The $1$-eigenspace of $S$ is spanned by vectors of the forms
(a) $e_i\otimes e_i\otimes e_i$ ;
(b) $e_i\otimes e_i\otimes e_j+ e_i\otimes e_j\otimes e_i+e_j\otimes e_i\otimes e_i$ ;
(c) $v_{(i,j,k)}:=e_i\otimes e_j\otimes e_k+ e_j\otimes e_k\otimes e_i+e_k\otimes e_i\otimes e_j$;
where the indices are distinct.
It can now be checked that $v_{(i,j,k)}-v_{(i,k,j)}$ is a common eigenvector of $T_1$ and $T_2$ both having eigenvalue $-1$.
(III) It would be good to list the triples $(a_1, a_2, w)$ with $T_i w= a_i w$ but this I have not done.
More Complete Answer (For those who know some character theory.)
What we have here is a representation of the group $\langle T_1, S=T_1 T_2\rangle \simeq \langle t, s \mid t^2=s^3=(ts)^2=1\rangle\simeq S_3$ on the space $W=V\otimes V\otimes V$ where $\dim V=n$. Call its character $\psi$.
The character table of $S_3$ is
$$
\begin{array}[l| c c c]
\ & 1 & s & t\\
\iota & 1 & 1 &1\\
\sigma & 1 & 1 & -1\\
\theta & 2& -1 & 0\\
\end{array}.
$$
We can easily calculate the traces $\psi(I)$, $\psi(S)$,  $\psi(T_1)$: as these matrices act by permuting the basis elements $e_i\otimes e_j\otimes e_k$ of $W$ we  need only count how many are fixed. We get at once that $\psi(I)=n^3$, $\psi(S)=n$, and $\psi(T_1)=n^2$. It is then easy to see how $\psi$ decomposes, we  must have
$$
\psi ={n+2 \choose  3}\iota +
{n \choose  3}\sigma +
2{n+1 \choose  3}\theta.
$$
That means that $W$ decomposes into ${n+2 \choose  3}$ $1$ dimensional subspaces on which both $T_1$ and $T_2$ act as multiplication by $+1$; ${n \choose  3}$ $1$ dimensional subspaces on which both $T_1$ and $T_2$ act as as multiplication by $+1$; and $2{n+1 \choose  3}$ $2$ dimensional subspaces where there are no common eigenvectors.
