= Solution
The two diagonal $2\times2$ blocks allow the <characteristic polynomial> to factor immediately:
$$
\det(\lambda I-A)=\bigl[(\lambda-i)^2-1\bigr]\bigl[(\lambda-i)^2+1\bigr].
$$
Thus the <characteristic equation> is
$$
\boxed{(\lambda-i-1)(\lambda-i+1)\lambda(\lambda-2i)=0.}
$$
Solving the two block <eigenvalue> equations gives the following nonzero <eigenvectors> and corresponding <eigenvalues>:
$$
\begin{array}{c|c}
\text{eigenvector}&\text{eigenvalue}\\\hline
a=(1,1,0,0)^T&1+i\\
b=(1,-1,0,0)^T&-1+i\\
c=(0,0,1,i)^T&2i\\
d=(0,0,1,-i)^T&0
\end{array}
$$
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 $(-2)(-2i)=4i\ne0$. Hence they form a <basis> and \b[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 $A(sv)=\lambda_vsv$. If $\lambda_v\ne0$, every $tv$ has preimage $(t/\lambda_v)v$, so the restricted <linear map> is onto that same subspace. Consequently
$$
\boxed{A(\mathbb Ca)=\mathbb Ca,\quad A(\mathbb Cb)=\mathbb Cb,\quad A(\mathbb Cc)=\mathbb Cc,\quad A(\mathbb Cd)=\{0\}.}
$$
The last line is the <kernel of a linear map> direction: multiplication by $A$ collapses that line to the zero-dimensional subspace.
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