 Solution

The two diagonal $2\times 2$ blocks allow the characteristic polynomial to factor immediately:Thus the characteristic equation isSolving the two block eigenvalue equations gives the following nonzero eigenvectors and corresponding eigenvalues:For example, the lower block maps $(1,i{)}^T$ to $(2i,-2{)}^T=2i(1,i{)}^T$, and maps $(1,-i{)}^T$ to zero. The four eigenvalues are distinct. Alternatively, the determinant of the matrix with columns $a,b,c,d$ is $(-2)(-2i)=4i\ne 0$. Hence they form a basis and span the complex vector space ${\mathbb{C}}^4$. Their nonzero scalar multiples are all the corresponding eigenvectors.

For any eigenvector $v$ with eigenvalue ${\lambda }_v$, its one-dimensional vector subspace satisfies $A(sv)={\lambda }_{v}sv$. If ${\lambda }_v\ne 0$, every $tv$ has preimage $(t/{\lambda }_v)v$, so the restricted linear map is onto that same subspace. ConsequentlyThe last line is the kernel of a linear map direction: multiplication by $A$ collapses that line to the zero-dimensional subspace.