# 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

## 3 Answers

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

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

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

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

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

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

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.