Eigenvectors and matrix decomposition of a Quaternion

Question

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 $Q$

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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?)

Answer 1

$$\det(tI-A)= \begin{vmatrix}t-d&c&\!\!-b&-a\\\!\!-c&t-d&a&-b\\b&\!\!-a&t-d&-c\\a&b&c&t-d\end{vmatrix}=$$

$$(t-d) \begin{vmatrix}t-d&a&-b\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+c \begin{vmatrix}c&\!\!-b&-a\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+b\begin{vmatrix}c&\!\!-b&-a\\t-d&a&-b\\b&c&t-d\end{vmatrix}-$$$${}$$

$$-a\begin{vmatrix}c&\!\!-b&-a\\t-d&a&-b\\\!\!-a&t-d&-c\end{vmatrix}=(t-d)^2\left[(t-d)^2+a^2+b^2+^2\right]+$$$${}$$

$$+c^2\left[(t-d)^2+a^2+b^2+c^2\right]+b^2\left[a^2+b^2+c^2+(t-d)^2\right]-$$$${}$$

$$-a^2\left[-a^2-b^2-c^2-(t-d)^2\right]=\left[(t-d)^2+a^2+b^2+c^2\right]^2$$$${}$$

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.

$$$$

Answer 2

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.

Answer 3

For a (4D) geometrical interpretation, it is useful to think of quaternion multiplications as scalings of special orthogonal operators.

Each of the eigenvalues of a quaternion multiplication operator is of multiplicity 2.

The eigenspace corresponding to a real eigenvalue is a 2-dimensional plane in 4-space.

The real and imaginary parts of the eigenvectors of a complex pair of eigenvalues span a 2-dimensional plane in 4-space.

In each case, the 2-dimensional planes are invariant under quaternion multiplication. (That is, the image of a vector in the plane under the multiplication is again in the plane.)

Within the planes corresponding to real eigenvalues, points are just scaled under quaternion multiplication. Within the planes corresponding to complex eigenvalues, vectors are scaled and rotated.

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.

Multiplication by a non-real quaternion always rotates points in its two invariant planes by the same amount. So the size of the moduli of the complex eigenvalues of a quaternion multiplication are equal to one another.

That's just quaternion multiplication on one side or the other, though. In general, special orthogonal operators of 4-space are a product of a left and a right quaternion multiplication (for a total of 6 degrees of freedom), and they do not typically have any invariant planes. But when they have one invariant plane, the perpendicular plane is also invariant.

When a multiplication on the left by a quaternion is combined with a multiplication on the right by its inverse, the result leaves points in one invariant plane stationary, but rotates points in the other invariant plane. This is a 3D rotation in a 3D subspace 4-space. It has one eigenvalue equal to 1 whose eigenvectors are in the plane containing the axis, and one complex eigenvalue, corresponding the the rotation in the other plane.