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?)
 A: $$\det(tI-A)= \begin{vmatrix}t-d&c&\!\!-b&-a\\\!\!-c&t-d&a&-b\\b&\!\!-a&t-d&-c\\a&b&c&t-d\end{vmatrix}=$$
$$(t-d) \begin{vmatrix}t-d&a&-b\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+c \begin{vmatrix}c&\!\!-b&-a\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+b\begin{vmatrix}c&\!\!-b&-a\\t-d&a&-b\\b&c&t-d\end{vmatrix}-$$$${}$$
$$-a\begin{vmatrix}c&\!\!-b&-a\\t-d&a&-b\\\!\!-a&t-d&-c\end{vmatrix}=(t-d)^2\left[(t-d)^2+a^2+b^2+^2\right]+$$$${}$$
$$+c^2\left[(t-d)^2+a^2+b^2+c^2\right]+b^2\left[a^2+b^2+c^2+(t-d)^2\right]-$$$${}$$
$$-a^2\left[-a^2-b^2-c^2-(t-d)^2\right]=\left[(t-d)^2+a^2+b^2+c^2\right]^2$$$${}$$
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.
$$$$
A: 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.
A: I spent a long time wondering about this myself.  Here are some metamathematical thoughts.
The idea of eigenanalysis is to find a way in which the action of a linear transform behaves like multiplication by a scalar, in the hope of simplifying the action of the transformation.  For many transforms with real eigenvalues, this has neat interpretations.  For a 2D rotation however, there isn't any neat interpretation as multiplication by real numbers.  The easy extension of the idea uses complex eigenvalues -- whereby the rotation is exactly represented by complex multiplication by that eigenvalue, which itself has a geometric interpretation as the same rotation.
The fundamental theorem of algebra is equivalent to the existence of a complex eigenvalue of any mapping of complex n-space to itself.  That is very neat.  But is it the answer to all questions in linear algebra?  (No, it's not.)
One question that might be asked here is: to what degree is an eigenanalysis of a quaternion useful?  Another might be, what does an eigenanalysis even mean?
You're asking for eigenvalues of a transform equivalent to (left- or right-) multiplication by a quaternion.  What exactly do you hope for?  Well, real eigenvalues are usually too much to expect, given what we know of general real transformations.  So ... could it be that a quaternion behaves like complex multiplication with some other quaternion?  Well, it could.  Besides quaternions that are simple scalings (which have a single real eigenvalue), some quaternions represent simple rotations of 3-space, and thus have complex and real eigenvalues.  But not all quaternions behave so: typically they model a richer set of changes of orientation (a rotation and a twist).
But the whole point of the construction of the quaternions is to produce a four-dimensional entity that behaves somehow as a scalar.  What is a "scalar" though?  It's a matter of semantics: if the term means something that is only a "size", it rules out complex scalars.  In recent decades the term has been extended to include the complex numbers, and specifically meant to exclude anything else -- specifically to accommodate things like eigenanalysis.  But if the term means an element of an algebra that behaves somehow like numbers, say, element of a division ring, and you're not picky about commutativity, quaternions can be viewed as a sort of algebra of scalars.
Like the situation with eigenvalues of a complex number, the most satisfying interpretation of an "eigenvalue" for a typical quaternion may be: the quaternion itself.
A: A (4D) geometrical interpretation is this.
Each of your eigenvalues is of multiplicity 2.  The eigenspace corresponding to each eigenvalue is a 2-dimensional plane in 4-space.
Multiplication by a quaternion is an orthogonal transformation of 4-space which rotates each of these planes by the argument of the corresponding eigenvector.  The two arguments are not independent for a given quaternion, however.
That's just quaternion multiplication on one side or the other, though.  General orthogonal transformations of 4-space are a product of a left and a right quaternion multiplication (for a total of 6 degrees of freedom).
Yes, orthogonal transformations of 4-space can consist of two independent analogs of rotation, but with the axis of rotation of each being a plane.
