Eigenvectors and matrix decomposition of a Quaternion Given the matrix representation of Quaternions
(re. e.g. to this other post)
$$
Q \ :=  \ \left(\begin{array}{rrrr}d&-c&b&a\\c&d&-a&b\\-b&a&d&c\\-a&-b&-c&d\end{array}\right) \ \
$$
what "meaning" or "role" can be given to the eigenvectors? and what to the decompositions of $Q$

p.s.
The eigenvalues result to be
$d \pm \sqrt { - \left( {a^2  + b^2  + c^2 } \right)} $
each with multiplicity $2$
and the eigenvectors
$$
\left( {\begin{array}{*{20}c}
   { - bq - ac} & { - aq + bc} & {bq - ac} & {aq + bc}  \\
   {aq - bc} & { - bq - ac} & { - aq - bc} & {bq - ac}  \\
   {a^2  + b^2 } & 0 & {a^2  + b^2 } & 0  \\
   0 & {a^2  + b^2 } & 0 & {a^2  + b^2 }  \\
 \end{array} } \right)\quad \left| {\;q = \sqrt { - \left( {a^2  + b^2  + c^2 } \right)} } \right.
$$
p.s. 2
Following @greg's answer, if $q$ could be "accomodated in", then the matrix would be diagonalizable, and powers and Taylor series easily computable ... .
So my question translates into whether such "accomodation" is fully out of quaternions algebra (-> e.g. the exp(Q) calculated through diagonalization is meaningful?)

Given a quaternion $q=ai+bj+ck+d\in\mathbb{H}$, we can consider the $\mathbb{R}$-linear map on the quaternions given by $w\mapsto qw$. The matrix of this linear map with respect to the basis $\{i,j,k,1\}$ is exactly $Q$. Thus a real eigenvalue of $Q$ should be a real number $\lambda$ such that there exists $w\in\mathbb{H}$, $w\ne0$, with $qw=\lambda w$, which can obviously happen only when $q=\lambda$, that is, $a=b=c=0$.
Since there is no "good" embedding of the complex numbers in the quaternions (there are infinitely many of them), there's no particular way for interpreting complex eigenvectors in this context.
