If $$A = \left[ {\matrix{ 1 & 0 & 0 & 1 \cr 0 & { - 1} & 0 & { - 1} \cr 0 & 0 & i & i \cr 0 & 0 & 0 & { - i} \cr } } \right]$$ the matrix $${A^4},$$
calculated by the use of Cayley - Hamilton theoram (or) otherwise is

The eigen vector (s) of the matrix
$$\left[ {\matrix{ 0 & 0 & \alpha \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right],\alpha \ne 0$$ is (are)