Is there an example for the following problem? Find a matrix $M$ with two Jordan blocks $J_1(\lambda), J_2(\lambda)$ of the same size to the same eigenvalue $\lambda$ such that the following holds:
Let $v_1$ be the right eigenvector to $J_1(\lambda)$ and $v_2$ be the right eigenvector to $J_2(\lambda)$ and let $u_1, u_2$ be the respective dual eigenvectors (i.e. the eigenvectors of $M^T$ to the respective blocks with $u_i^T v_j = \delta_{i,j}$).
Then for any combination of entries $(i,j)$ one has $u_{1,i} v_{1,j} = - u_{2,i} v_{2,j}$.
Ok I found a solution to my question. The answer is that such a thing does not exist. The question is equivalent to the following: For all matrices $B$ with one entry $1$ and all other entries $0$ $$ u_1 B v_1 + u_2 B v_2 = 0. $$ Now each $u_k B v_k$ is a scalar hence we can apply vectorization and the tensor-vectorization identity to it to obtain \begin{equation} \begin{split} u_1 B v_1 + u_2 B v_2 &= \mathsf{vec}(u_1 B v_1) + \mathsf{vec}(u_2 B v_2) \\ &= (v_1 \otimes u_1) \mathsf{vec}(B) + (v_2 \otimes u_2) \mathsf{vec}(B) \\ &= ((v_1 \otimes u_1) + (v_2 \otimes u_2)) \mathsf{vec}(B) \end{split} \end{equation} The vectors $u_k, v_k$ are elements of bases, so are $v_k \otimes u_k$. Thus there must be atleast one entry of $(v_1 \otimes u_1) + (v_2 \otimes u_2)$ which is non-zero, which can be singled out with one of the above matrice $B$.