# 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?)

• Did you compute the eigenvalues? – quid Sep 7 '16 at 10:56
• Yes, I did, but can not figure out yet if joined up they give a quaternion or not. – G Cab Sep 7 '16 at 11:05
• Alright. So maybe include this information in your question. – quid Sep 7 '16 at 11:05
• @DonAntonio It's the matrix corresponding to the multiplication by $ai+bi+cj+d$ with respect to the basis $\{i,j,k,1\}$. – egreg Sep 7 '16 at 11:30
• @quid, for better reference I included the computed eigenvalues and eigenvectors, which in fact are not real – G Cab Sep 7 '16 at 13:00

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

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

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

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

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

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.



• Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ? – G Cab Sep 7 '16 at 13:18

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.

• Thanks, good to know that there is no "good" embedding of complex numbers in quaternions: that's a pity because the eigenvalues look to have a "nice" expression. – G Cab Sep 7 '16 at 13:09