When do two linear hermitian operators have a common eigen vector? Let $H$ be an finite dimension hilbert space. Let $L_1$ and $L_2$ be two hermitian linear operators acting on this space. I know if these two operators commute they can be diagonalized in a common orthonormal basis. Thus they have common eigen vectors. But if they don't commute can they have a common eigen vector. I might be wrong but is commuting the necessary and sufficient condition for two hermitian operators to have a common eigen vector ?
 A: No. The matrices 
\begin{align*}
A &=\begin{bmatrix}5/3&0&-2/3\\0&7/3&-2/3\\-2/3&-2/3&2\end{bmatrix} & 
B &=\begin{bmatrix}37/18&-2/9&-17/18\\ -2/9&17/9&-2/9\\-17/18&-2/9&37/18\end{bmatrix}
\end{align*}
both have $\vec v=\begin{bmatrix}2\\1\\2\end{bmatrix}$ as an eigenvector and corresponding eigenvalue $\lambda=1$. However
$$
AB=\begin{bmatrix}73/18&-2/9&-53/18\\ 1/9&41/9&-17/9\\-28/9&-14/9&44/9\end{bmatrix}\neq
\begin{bmatrix}
73/18&1/9&-28/9\\-2/9&41/9&-14/9\\-53/18&-17/9&44/9
\end{bmatrix}=BA
$$
To generate more counterexamples, let $\vec u$ be a unit vector and extend $\vec u$ to two different orthonormal bases of $\Bbb R^n$. Put these two bases into the columns of matrices $P$ and $Q$. Let $D$ and $E$ be diagonal matrices and put
\begin{align*}
A &= PDP^\top & B &= QEQ^\top
\end{align*}
Then $A$ and $B$ are symmetric matrices that share an eigenvalue $\vec u$ but $AB\neq BA$ in general.
A: If two diagonalizable linear operators commute then ALL of their eigenvectors are the same, and at least in the finite dimensional case this is if and only if. Thus you can choose one eigenvector $v$ and extend to two different bases $V_1$ and $V_2$ of eigenvectors, and choose whatever distinct eigenvalues you want and you will have two linear operators with exactly one common eigenvector that do not commute.
