Mixing
Eigenvectors and matrix decomposition of a Quaternion
$begingroup$
Given the matrix representation of Quaternions
(re. e.g. to this other post)
$$
Q := left(begin{array}{rrrr}d&-c&b&a\c&d&-a&b\-b&a&d&c\-a&-b&-c&dend{array}right)
$$
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?)
quaternions
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
$endgroup$
|
show 3 more comments
$begingroup$
Given the matrix representation of Quaternions
(re. e.g. to this other post)
$$
Q := left(begin{array}{rrrr}d&-c&b&a\c&d&-a&b\-b&a&d&c\-a&-b&-c&dend{array}right)
$$
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?)
quaternions
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
$endgroup$
1
$begingroup$
Did you compute the eigenvalues?
$endgroup$
– quid♦
Sep 7 '16 at 10:56
$begingroup$
Yes, I did, but can not figure out yet if joined up they give a quaternion or not.
$endgroup$
– G Cab
Sep 7 '16 at 11:05
1
$begingroup$
Alright. So maybe include this information in your question.
$endgroup$
– quid♦
Sep 7 '16 at 11:05
2
$begingroup$
@DonAntonio It's the matrix corresponding to the multiplication by $ai+bi+cj+d$ with respect to the basis ${i,j,k,1}$.
$endgroup$
– egreg
Sep 7 '16 at 11:30
1
$begingroup$
@quid, for better reference I included the computed eigenvalues and eigenvectors, which in fact are not real
$endgroup$
– G Cab
Sep 7 '16 at 13:00
|
show 3 more comments
$begingroup$
Given the matrix representation of Quaternions
(re. e.g. to this other post)
$$
Q := left(begin{array}{rrrr}d&-c&b&a\c&d&-a&b\-b&a&d&c\-a&-b&-c&dend{array}right)
$$
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?)
quaternions
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
$endgroup$
Given the matrix representation of Quaternions
(re. e.g. to this other post)
$$
Q := left(begin{array}{rrrr}d&-c&b&a\c&d&-a&b\-b&a&d&c\-a&-b&-c&dend{array}right)
$$
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?)
quaternions
quaternions
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
asked Sep 7 '16 at 10:54
G CabG Cab
18.9k31238
18.9k31238
1
$begingroup$
Did you compute the eigenvalues?
$endgroup$
– quid♦
Sep 7 '16 at 10:56
$begingroup$
Yes, I did, but can not figure out yet if joined up they give a quaternion or not.
$endgroup$
– G Cab
Sep 7 '16 at 11:05
1
$begingroup$
Alright. So maybe include this information in your question.
$endgroup$
– quid♦
Sep 7 '16 at 11:05
2
$begingroup$
@DonAntonio It's the matrix corresponding to the multiplication by $ai+bi+cj+d$ with respect to the basis ${i,j,k,1}$.
$endgroup$
– egreg
Sep 7 '16 at 11:30
1
$begingroup$
@quid, for better reference I included the computed eigenvalues and eigenvectors, which in fact are not real
$endgroup$
– G Cab
Sep 7 '16 at 13:00
|
show 3 more comments
1
$begingroup$
Did you compute the eigenvalues?
$endgroup$
– quid♦
Sep 7 '16 at 10:56
$begingroup$
Yes, I did, but can not figure out yet if joined up they give a quaternion or not.
$endgroup$
– G Cab
Sep 7 '16 at 11:05
1
$begingroup$
Alright. So maybe include this information in your question.
$endgroup$
– quid♦
Sep 7 '16 at 11:05
2
$begingroup$
@DonAntonio It's the matrix corresponding to the multiplication by $ai+bi+cj+d$ with respect to the basis ${i,j,k,1}$.
$endgroup$
– egreg
Sep 7 '16 at 11:30
1
$begingroup$
@quid, for better reference I included the computed eigenvalues and eigenvectors, which in fact are not real
$endgroup$
– G Cab
Sep 7 '16 at 13:00
1
1
$begingroup$
Did you compute the eigenvalues?
$endgroup$
– quid♦
Sep 7 '16 at 10:56
$begingroup$
Did you compute the eigenvalues?
$endgroup$
– quid♦
Sep 7 '16 at 10:56
$begingroup$
Yes, I did, but can not figure out yet if joined up they give a quaternion or not.
$endgroup$
– G Cab
Sep 7 '16 at 11:05
$begingroup$
Yes, I did, but can not figure out yet if joined up they give a quaternion or not.
$endgroup$
– G Cab
Sep 7 '16 at 11:05
1
1
$begingroup$
Alright. So maybe include this information in your question.
$endgroup$
– quid♦
Sep 7 '16 at 11:05
$begingroup$
Alright. So maybe include this information in your question.
$endgroup$
– quid♦
Sep 7 '16 at 11:05
2
2
$begingroup$
@DonAntonio It's the matrix corresponding to the multiplication by $ai+bi+cj+d$ with respect to the basis ${i,j,k,1}$.
$endgroup$
– egreg
Sep 7 '16 at 11:30
$begingroup$
@DonAntonio It's the matrix corresponding to the multiplication by $ai+bi+cj+d$ with respect to the basis ${i,j,k,1}$.
$endgroup$
– egreg
Sep 7 '16 at 11:30
1
1
$begingroup$
@quid, for better reference I included the computed eigenvalues and eigenvectors, which in fact are not real
$endgroup$
– G Cab
Sep 7 '16 at 13:00
$begingroup$
@quid, for better reference I included the computed eigenvalues and eigenvectors, which in fact are not real
$endgroup$
– G Cab
Sep 7 '16 at 13:00
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
$$det(tI-A)= begin{vmatrix}t-d&c&!!-b&-a\!!-c&t-d&a&-b\b&!!-a&t-d&-c\a&b&c&t-dend{vmatrix}=$$
$$(t-d) begin{vmatrix}t-d&a&-b\!!-a&t-d&-c\b&c&t-dend{vmatrix}+c begin{vmatrix}c&!!-b&-a\!!-a&t-d&-c\b&c&t-dend{vmatrix}+bbegin{vmatrix}c&!!-b&-a\t-d&a&-b\b&c&t-dend{vmatrix}-$$$${}$$
$$-abegin{vmatrix}c&!!-b&-a\t-d&a&-b\!!-a&t-d&-cend{vmatrix}=(t-d)^2left[(t-d)^2+a^2+b^2+^2right]+$$$${}$$
$$+c^2left[(t-d)^2+a^2+b^2+c^2right]+b^2left[a^2+b^2+c^2+(t-d)^2right]-$$$${}$$
$$-a^2left[-a^2-b^2-c^2-(t-d)^2right]=left[(t-d)^2+a^2+b^2+c^2right]^2$$$${}$$
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.
$$$$
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
$endgroup$
$begingroup$
Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ?
$endgroup$
– G Cab
Sep 7 '16 at 13:18
add a comment |
$begingroup$
Given a quaternion $q=ai+bj+ck+dinmathbb{H}$, we can consider the $mathbb{R}$-linear map on the quaternions given by $wmapsto 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 $winmathbb{H}$, $wne0$, 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.
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
$endgroup$
$begingroup$
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.
$endgroup$
– G Cab
Sep 7 '16 at 13:09
add a comment |
$begingroup$
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.
answered Jan 9 at 11:44
Steve WhiteSteve White
263
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$det(tI-A)= begin{vmatrix}t-d&c&!!-b&-a\!!-c&t-d&a&-b\b&!!-a&t-d&-c\a&b&c&t-dend{vmatrix}=$$
$$(t-d) begin{vmatrix}t-d&a&-b\!!-a&t-d&-c\b&c&t-dend{vmatrix}+c begin{vmatrix}c&!!-b&-a\!!-a&t-d&-c\b&c&t-dend{vmatrix}+bbegin{vmatrix}c&!!-b&-a\t-d&a&-b\b&c&t-dend{vmatrix}-$$$${}$$
$$-abegin{vmatrix}c&!!-b&-a\t-d&a&-b\!!-a&t-d&-cend{vmatrix}=(t-d)^2left[(t-d)^2+a^2+b^2+^2right]+$$$${}$$
$$+c^2left[(t-d)^2+a^2+b^2+c^2right]+b^2left[a^2+b^2+c^2+(t-d)^2right]-$$$${}$$
$$-a^2left[-a^2-b^2-c^2-(t-d)^2right]=left[(t-d)^2+a^2+b^2+c^2right]^2$$$${}$$
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.
$$$$
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
$endgroup$
$begingroup$
Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ?
$endgroup$
– G Cab
Sep 7 '16 at 13:18
add a comment |
$begingroup$
$$det(tI-A)= begin{vmatrix}t-d&c&!!-b&-a\!!-c&t-d&a&-b\b&!!-a&t-d&-c\a&b&c&t-dend{vmatrix}=$$
$$(t-d) begin{vmatrix}t-d&a&-b\!!-a&t-d&-c\b&c&t-dend{vmatrix}+c begin{vmatrix}c&!!-b&-a\!!-a&t-d&-c\b&c&t-dend{vmatrix}+bbegin{vmatrix}c&!!-b&-a\t-d&a&-b\b&c&t-dend{vmatrix}-$$$${}$$
$$-abegin{vmatrix}c&!!-b&-a\t-d&a&-b\!!-a&t-d&-cend{vmatrix}=(t-d)^2left[(t-d)^2+a^2+b^2+^2right]+$$$${}$$
$$+c^2left[(t-d)^2+a^2+b^2+c^2right]+b^2left[a^2+b^2+c^2+(t-d)^2right]-$$$${}$$
$$-a^2left[-a^2-b^2-c^2-(t-d)^2right]=left[(t-d)^2+a^2+b^2+c^2right]^2$$$${}$$
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.
$$$$
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
$endgroup$
$begingroup$
Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ?
$endgroup$
– G Cab
Sep 7 '16 at 13:18
add a comment |
$begingroup$
$$det(tI-A)= begin{vmatrix}t-d&c&!!-b&-a\!!-c&t-d&a&-b\b&!!-a&t-d&-c\a&b&c&t-dend{vmatrix}=$$
$$(t-d) begin{vmatrix}t-d&a&-b\!!-a&t-d&-c\b&c&t-dend{vmatrix}+c begin{vmatrix}c&!!-b&-a\!!-a&t-d&-c\b&c&t-dend{vmatrix}+bbegin{vmatrix}c&!!-b&-a\t-d&a&-b\b&c&t-dend{vmatrix}-$$$${}$$
$$-abegin{vmatrix}c&!!-b&-a\t-d&a&-b\!!-a&t-d&-cend{vmatrix}=(t-d)^2left[(t-d)^2+a^2+b^2+^2right]+$$$${}$$
$$+c^2left[(t-d)^2+a^2+b^2+c^2right]+b^2left[a^2+b^2+c^2+(t-d)^2right]-$$$${}$$
$$-a^2left[-a^2-b^2-c^2-(t-d)^2right]=left[(t-d)^2+a^2+b^2+c^2right]^2$$$${}$$
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.
$$$$
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
$endgroup$
$$det(tI-A)= begin{vmatrix}t-d&c&!!-b&-a\!!-c&t-d&a&-b\b&!!-a&t-d&-c\a&b&c&t-dend{vmatrix}=$$
$$(t-d) begin{vmatrix}t-d&a&-b\!!-a&t-d&-c\b&c&t-dend{vmatrix}+c begin{vmatrix}c&!!-b&-a\!!-a&t-d&-c\b&c&t-dend{vmatrix}+bbegin{vmatrix}c&!!-b&-a\t-d&a&-b\b&c&t-dend{vmatrix}-$$$${}$$
$$-abegin{vmatrix}c&!!-b&-a\t-d&a&-b\!!-a&t-d&-cend{vmatrix}=(t-d)^2left[(t-d)^2+a^2+b^2+^2right]+$$$${}$$
$$+c^2left[(t-d)^2+a^2+b^2+c^2right]+b^2left[a^2+b^2+c^2+(t-d)^2right]-$$$${}$$
$$-a^2left[-a^2-b^2-c^2-(t-d)^2right]=left[(t-d)^2+a^2+b^2+c^2right]^2$$$${}$$
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.
$$$$
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
answered Sep 7 '16 at 11:34
DonAntonioDonAntonio
178k1493228
178k1493228
$begingroup$
Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ?
$endgroup$
– G Cab
Sep 7 '16 at 13:18
add a comment |
$begingroup$
Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ?
$endgroup$
– G Cab
Sep 7 '16 at 13:18
$begingroup$
Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ?
$endgroup$
– G Cab
Sep 7 '16 at 13:18
$begingroup$
Thanks, and in fact they are complex, and I presume they are so also in "other" representations, is that correct ?
$endgroup$
– G Cab
Sep 7 '16 at 13:18
add a comment |
$begingroup$
Given a quaternion $q=ai+bj+ck+dinmathbb{H}$, we can consider the $mathbb{R}$-linear map on the quaternions given by $wmapsto qw$. 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 $winmathbb{H}$, $wne0$, 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.
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
$endgroup$
$begingroup$
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.
$endgroup$
– G Cab
Sep 7 '16 at 13:09
add a comment |
$begingroup$
Given a quaternion $q=ai+bj+ck+dinmathbb{H}$, we can consider the $mathbb{R}$-linear map on the quaternions given by $wmapsto qw$. 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 $winmathbb{H}$, $wne0$, 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.
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
$endgroup$
$begingroup$
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.
$endgroup$
– G Cab
Sep 7 '16 at 13:09
add a comment |
$begingroup$
Given a quaternion $q=ai+bj+ck+dinmathbb{H}$, we can consider the $mathbb{R}$-linear map on the quaternions given by $wmapsto qw$. 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 $winmathbb{H}$, $wne0$, 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.
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
$endgroup$
Given a quaternion $q=ai+bj+ck+dinmathbb{H}$, we can consider the $mathbb{R}$-linear map on the quaternions given by $wmapsto qw$. 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 $winmathbb{H}$, $wne0$, 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.
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
answered Sep 7 '16 at 12:29
egregegreg
181k1485203
181k1485203
$begingroup$
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.
$endgroup$
– G Cab
Sep 7 '16 at 13:09
add a comment |
$begingroup$
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.
$endgroup$
– G Cab
Sep 7 '16 at 13:09
$begingroup$
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.
$endgroup$
– G Cab
Sep 7 '16 at 13:09
$begingroup$
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.
$endgroup$
– G Cab
Sep 7 '16 at 13:09
add a comment |
$begingroup$
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.
answered Jan 9 at 11:44
Steve WhiteSteve White
263
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 9 at 11:44
Steve WhiteSteve White
263
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 9 at 11:44
Steve WhiteSteve White
263
$endgroup$
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.
answered Jan 9 at 11:44
Steve WhiteSteve White
263
answered Jan 9 at 11:44
Steve WhiteSteve White
263
answered Jan 9 at 11:44
Steve WhiteSteve White
263
answered Jan 9 at 11:44
Steve WhiteSteve White
263
263
add a comment |
add a comment |
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1
$begingroup$
Did you compute the eigenvalues?
$endgroup$
– quid♦
Sep 7 '16 at 10:56
$begingroup$
Yes, I did, but can not figure out yet if joined up they give a quaternion or not.
$endgroup$
– G Cab
Sep 7 '16 at 11:05
1
$begingroup$
Alright. So maybe include this information in your question.
$endgroup$
– quid♦
Sep 7 '16 at 11:05
2
$begingroup$
@DonAntonio It's the matrix corresponding to the multiplication by $ai+bi+cj+d$ with respect to the basis ${i,j,k,1}$.
$endgroup$
– egreg
Sep 7 '16 at 11:30
1
$begingroup$
@quid, for better reference I included the computed eigenvalues and eigenvectors, which in fact are not real
$endgroup$
– G Cab
Sep 7 '16 at 13:00