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?)

$$\det(tI-A)= \begin{vmatrix}t-d&c&\!\!-b&-a\\\!\!-c&t-d&a&-b\\b&\!\!-a&t-d&-c\\a&b&c&t-d\end{vmatrix}=$$
$$(t-d) \begin{vmatrix}t-d&a&-b\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+c \begin{vmatrix}c&\!\!-b&-a\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+b\begin{vmatrix}c&\!\!-b&-a\\t-d&a&-b\\b&c&t-d\end{vmatrix}-$$$${}$$
$$-a\begin{vmatrix}c&\!\!-b&-a\\t-d&a&-b\\\!\!-a&t-d&-c\end{vmatrix}=(t-d)^2\left[(t-d)^2+a^2+b^2+^2\right]+$$$${}$$
$$+c^2\left[(t-d)^2+a^2+b^2+c^2\right]+b^2\left[a^2+b^2+c^2+(t-d)^2\right]-$$$${}$$
$$-a^2\left[-a^2-b^2-c^2-(t-d)^2\right]=\left[(t-d)^2+a^2+b^2+c^2\right]^2$$$${}$$
Thus the eigenvalues aren't real ( except in the extreme case when $\;a=b=c=0\;$) , which doesn't surprise as the above matrix representation of quaternions is skew-symmetric.
$$$$
