Let $A$ be diagonalisable linear operator on $\mathbb{C}^n$, and let $v_1,\dots,v_n$ be its basis of eigenvectors equipped wit total order $\geq$. Furthermore, let $B$ be nilpotent linear operator which satisfies for each $i$ $$ Bv_i\geq v_i \quad \text{or}\quad Bv_i=0 $$ (so the matrix of $B$ is strictly upper triangular in the basis $\{v_1,\dots,v_n\}$).
Now let us denote by $C$ the linear operator $C=A+B$. Since $B$ is strictly upper triangular in our basis then $A$ and $C$ have precisely the same eigenvalues.
Let $u_\alpha$ and $v_\alpha$ be eigenvectors of $C$ and $A$ with the eigenvalue $\alpha\in\mathbb{C}$. Then I claim that $$ u_\alpha= c\ v_\alpha+\sum_{i} c_i\ v_i $$ where sum is over all vectors $v_i$ such that $v_i\geq v_\alpha$ and not all $c,c_i$ are zero.
My linear algebra is very rusty. I have checked the claim in many cases, but I can not find exact proof.