So, where $\lambda$ is the eigenvalue, and $A$ is the transform matrix:
$\displaystyle AX=\lambda IX$
$\displaystyle (A-\lambda I)X = 0$
Since X can only be 0, we know that $(A-\lambda I)$ must be rank 4 and therefore $(A-\lambda I)$ is row equivalent to $I$. Therefore you need $A$ to be some matrix that stays rank 4 no matter what you add/subtract from the diagonals.