Modifying unitary matrix eigenvalues by right multiplication by orthogonal matrix
$begingroup$
I have a matrix $U in U(n)$ ($U^* U=Id$), with eigenvalues $lambda_1, dots lambda_n in S^1$. I would like to know if its always possible to find a matrix $O in O(n)$ such that the eigenvalues $lambda^{'}_1, dots, lambda^{'}_n$ of $UO in U(n)$ can be written as:
$$lambda^{'}_j=e^{i pi alpha_j}$$
Where $alpha_j in [0,1)$.
There is an easy case. If $U$ is a diagonal matrix and $J subset {1, dots , n }$ is the subset such that for $j in J$ we have $lambda_j=x_j + iy_j$ with $x,y in mathbb{R}$ and $(y<0 vee x=-1)$. Then choosing:
$${O}_{mn}=begin{cases}
0 & mne n \
-1 & m=nin J \
1 & else
end{cases}$$
This will define an orthogonal matrix and $UO$ will be as desired.
The problem is that when $U$ is not a diagonal matrix the matrix $S=P^*OP$ for $P in U(n)$ and $O in O(n)$ is not generally an orthogonal matrix so there is no simple algorithm as far as I can see to find $O in O(n)$
matrices eigenvalues-eigenvectors lie-groups lie-algebras
asked Dec 26 '18 at 10:32
EladElad
499213
$endgroup$
add a comment |
$begingroup$
I have a matrix $U in U(n)$ ($U^* U=Id$), with eigenvalues $lambda_1, dots lambda_n in S^1$. I would like to know if its always possible to find a matrix $O in O(n)$ such that the eigenvalues $lambda^{'}_1, dots, lambda^{'}_n$ of $UO in U(n)$ can be written as:
$$lambda^{'}_j=e^{i pi alpha_j}$$
Where $alpha_j in [0,1)$.
There is an easy case. If $U$ is a diagonal matrix and $J subset {1, dots , n }$ is the subset such that for $j in J$ we have $lambda_j=x_j + iy_j$ with $x,y in mathbb{R}$ and $(y<0 vee x=-1)$. Then choosing:
$${O}_{mn}=begin{cases}
0 & mne n \
-1 & m=nin J \
1 & else
end{cases}$$
This will define an orthogonal matrix and $UO$ will be as desired.
The problem is that when $U$ is not a diagonal matrix the matrix $S=P^*OP$ for $P in U(n)$ and $O in O(n)$ is not generally an orthogonal matrix so there is no simple algorithm as far as I can see to find $O in O(n)$
matrices eigenvalues-eigenvectors lie-groups lie-algebras
asked Dec 26 '18 at 10:32
EladElad
499213
$endgroup$
$begingroup$
May I ask what is the motivation for this?
$endgroup$
– Torsten Schoeneberg
Dec 29 '18 at 19:38
1
$begingroup$
Sure. The unitary group acts transitively on Lagrangian subspaces with stabilizers orthogonal matrices. I want to create an invariant of a function to the Lagrangian grassmanian and for that to be well defined I need the eigenvalues to be “oriented “ in some way.
$endgroup$
– Elad
Dec 29 '18 at 21:51
1
$begingroup$
This invariant is some version of the Maslov index ncatlab.org/nlab/show/Maslov+index
$endgroup$
– Elad
Dec 29 '18 at 22:04
add a comment |
$begingroup$
I have a matrix $U in U(n)$ ($U^* U=Id$), with eigenvalues $lambda_1, dots lambda_n in S^1$. I would like to know if its always possible to find a matrix $O in O(n)$ such that the eigenvalues $lambda^{'}_1, dots, lambda^{'}_n$ of $UO in U(n)$ can be written as:
$$lambda^{'}_j=e^{i pi alpha_j}$$
Where $alpha_j in [0,1)$.
There is an easy case. If $U$ is a diagonal matrix and $J subset {1, dots , n }$ is the subset such that for $j in J$ we have $lambda_j=x_j + iy_j$ with $x,y in mathbb{R}$ and $(y<0 vee x=-1)$. Then choosing:
$${O}_{mn}=begin{cases}
0 & mne n \
-1 & m=nin J \
1 & else
end{cases}$$
This will define an orthogonal matrix and $UO$ will be as desired.
The problem is that when $U$ is not a diagonal matrix the matrix $S=P^*OP$ for $P in U(n)$ and $O in O(n)$ is not generally an orthogonal matrix so there is no simple algorithm as far as I can see to find $O in O(n)$
matrices eigenvalues-eigenvectors lie-groups lie-algebras
asked Dec 26 '18 at 10:32
EladElad
499213
$endgroup$
I have a matrix $U in U(n)$ ($U^* U=Id$), with eigenvalues $lambda_1, dots lambda_n in S^1$. I would like to know if its always possible to find a matrix $O in O(n)$ such that the eigenvalues $lambda^{'}_1, dots, lambda^{'}_n$ of $UO in U(n)$ can be written as:
$$lambda^{'}_j=e^{i pi alpha_j}$$
Where $alpha_j in [0,1)$.
There is an easy case. If $U$ is a diagonal matrix and $J subset {1, dots , n }$ is the subset such that for $j in J$ we have $lambda_j=x_j + iy_j$ with $x,y in mathbb{R}$ and $(y<0 vee x=-1)$. Then choosing:
$${O}_{mn}=begin{cases}
0 & mne n \
-1 & m=nin J \
1 & else
end{cases}$$
This will define an orthogonal matrix and $UO$ will be as desired.
The problem is that when $U$ is not a diagonal matrix the matrix $S=P^*OP$ for $P in U(n)$ and $O in O(n)$ is not generally an orthogonal matrix so there is no simple algorithm as far as I can see to find $O in O(n)$
matrices eigenvalues-eigenvectors lie-groups lie-algebras
matrices eigenvalues-eigenvectors lie-groups lie-algebras
asked Dec 26 '18 at 10:32
EladElad
499213
asked Dec 26 '18 at 10:32
EladElad
499213
edited Dec 26 '18 at 10:39
Elad
asked Dec 26 '18 at 10:32
EladElad
499213
asked Dec 26 '18 at 10:32
EladElad
499213
asked Dec 26 '18 at 10:32
EladElad
499213
499213
$begingroup$
May I ask what is the motivation for this?
$endgroup$
– Torsten Schoeneberg
Dec 29 '18 at 19:38
1
$begingroup$
Sure. The unitary group acts transitively on Lagrangian subspaces with stabilizers orthogonal matrices. I want to create an invariant of a function to the Lagrangian grassmanian and for that to be well defined I need the eigenvalues to be “oriented “ in some way.
$endgroup$
– Elad
Dec 29 '18 at 21:51
1
$begingroup$
This invariant is some version of the Maslov index ncatlab.org/nlab/show/Maslov+index
$endgroup$
– Elad
Dec 29 '18 at 22:04
add a comment |
$begingroup$
May I ask what is the motivation for this?
$endgroup$
– Torsten Schoeneberg
Dec 29 '18 at 19:38
1
$begingroup$
Sure. The unitary group acts transitively on Lagrangian subspaces with stabilizers orthogonal matrices. I want to create an invariant of a function to the Lagrangian grassmanian and for that to be well defined I need the eigenvalues to be “oriented “ in some way.
$endgroup$
– Elad
Dec 29 '18 at 21:51
1
$begingroup$
This invariant is some version of the Maslov index ncatlab.org/nlab/show/Maslov+index
$endgroup$
– Elad
Dec 29 '18 at 22:04
$begingroup$
May I ask what is the motivation for this?
$endgroup$
– Torsten Schoeneberg
Dec 29 '18 at 19:38
$begingroup$
May I ask what is the motivation for this?
$endgroup$
– Torsten Schoeneberg
Dec 29 '18 at 19:38
1
1
$begingroup$
Sure. The unitary group acts transitively on Lagrangian subspaces with stabilizers orthogonal matrices. I want to create an invariant of a function to the Lagrangian grassmanian and for that to be well defined I need the eigenvalues to be “oriented “ in some way.
$endgroup$
– Elad
Dec 29 '18 at 21:51
$begingroup$
Sure. The unitary group acts transitively on Lagrangian subspaces with stabilizers orthogonal matrices. I want to create an invariant of a function to the Lagrangian grassmanian and for that to be well defined I need the eigenvalues to be “oriented “ in some way.
$endgroup$
– Elad
Dec 29 '18 at 21:51
1
1
$begingroup$
This invariant is some version of the Maslov index ncatlab.org/nlab/show/Maslov+index
$endgroup$
– Elad
Dec 29 '18 at 22:04
$begingroup$
This invariant is some version of the Maslov index ncatlab.org/nlab/show/Maslov+index
$endgroup$
– Elad
Dec 29 '18 at 22:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not exactly what you asked but it might help to solve this or narrow the search for a counter example.
We have $U in U(n) implies spec(U)subset S^1$. $spec(U)$ is discrete and therfore any function on it is continuous.
Define $f:spec(U) to {-1,1}$
$$e^{ipi theta} mapsto
begin{cases}
1 & theta in [0,pi) \
-1 & theta in [pi,2pi)
end{cases}$$
Note that $bar{f}f=f^2=1$. Set $O=f(U)$.
Since $f$ has image in $mathbb{R}cap S^1$ then
$$O in U(n)cap H(n)$$
Moreover, for all $x in Spec(U)$:
$$overline{xf(x)}xf(x)=1$$
so $(UO)^*(UO)=I$
because the continuous functional calculus is a $C^*$ algebra homomrphisim.
Finally $xf(x) in S^1 cap mathbb{H}$ and therefor $UO$ has spectrum in $S^1 cap mathbb{H}$.
$U(n) cap H(n)$ has similar properties to the orthogonal group so this might help.
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
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add a comment |
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$begingroup$
This is not exactly what you asked but it might help to solve this or narrow the search for a counter example.
We have $U in U(n) implies spec(U)subset S^1$. $spec(U)$ is discrete and therfore any function on it is continuous.
Define $f:spec(U) to {-1,1}$
$$e^{ipi theta} mapsto
begin{cases}
1 & theta in [0,pi) \
-1 & theta in [pi,2pi)
end{cases}$$
Note that $bar{f}f=f^2=1$. Set $O=f(U)$.
Since $f$ has image in $mathbb{R}cap S^1$ then
$$O in U(n)cap H(n)$$
Moreover, for all $x in Spec(U)$:
$$overline{xf(x)}xf(x)=1$$
so $(UO)^*(UO)=I$
because the continuous functional calculus is a $C^*$ algebra homomrphisim.
Finally $xf(x) in S^1 cap mathbb{H}$ and therefor $UO$ has spectrum in $S^1 cap mathbb{H}$.
$U(n) cap H(n)$ has similar properties to the orthogonal group so this might help.
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
$endgroup$
add a comment |
$begingroup$
This is not exactly what you asked but it might help to solve this or narrow the search for a counter example.
We have $U in U(n) implies spec(U)subset S^1$. $spec(U)$ is discrete and therfore any function on it is continuous.
Define $f:spec(U) to {-1,1}$
$$e^{ipi theta} mapsto
begin{cases}
1 & theta in [0,pi) \
-1 & theta in [pi,2pi)
end{cases}$$
Note that $bar{f}f=f^2=1$. Set $O=f(U)$.
Since $f$ has image in $mathbb{R}cap S^1$ then
$$O in U(n)cap H(n)$$
Moreover, for all $x in Spec(U)$:
$$overline{xf(x)}xf(x)=1$$
so $(UO)^*(UO)=I$
because the continuous functional calculus is a $C^*$ algebra homomrphisim.
Finally $xf(x) in S^1 cap mathbb{H}$ and therefor $UO$ has spectrum in $S^1 cap mathbb{H}$.
$U(n) cap H(n)$ has similar properties to the orthogonal group so this might help.
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
$endgroup$
add a comment |
$begingroup$
This is not exactly what you asked but it might help to solve this or narrow the search for a counter example.
We have $U in U(n) implies spec(U)subset S^1$. $spec(U)$ is discrete and therfore any function on it is continuous.
Define $f:spec(U) to {-1,1}$
$$e^{ipi theta} mapsto
begin{cases}
1 & theta in [0,pi) \
-1 & theta in [pi,2pi)
end{cases}$$
Note that $bar{f}f=f^2=1$. Set $O=f(U)$.
Since $f$ has image in $mathbb{R}cap S^1$ then
$$O in U(n)cap H(n)$$
Moreover, for all $x in Spec(U)$:
$$overline{xf(x)}xf(x)=1$$
so $(UO)^*(UO)=I$
because the continuous functional calculus is a $C^*$ algebra homomrphisim.
Finally $xf(x) in S^1 cap mathbb{H}$ and therefor $UO$ has spectrum in $S^1 cap mathbb{H}$.
$U(n) cap H(n)$ has similar properties to the orthogonal group so this might help.
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
$endgroup$
This is not exactly what you asked but it might help to solve this or narrow the search for a counter example.
We have $U in U(n) implies spec(U)subset S^1$. $spec(U)$ is discrete and therfore any function on it is continuous.
Define $f:spec(U) to {-1,1}$
$$e^{ipi theta} mapsto
begin{cases}
1 & theta in [0,pi) \
-1 & theta in [pi,2pi)
end{cases}$$
Note that $bar{f}f=f^2=1$. Set $O=f(U)$.
Since $f$ has image in $mathbb{R}cap S^1$ then
$$O in U(n)cap H(n)$$
Moreover, for all $x in Spec(U)$:
$$overline{xf(x)}xf(x)=1$$
so $(UO)^*(UO)=I$
because the continuous functional calculus is a $C^*$ algebra homomrphisim.
Finally $xf(x) in S^1 cap mathbb{H}$ and therefor $UO$ has spectrum in $S^1 cap mathbb{H}$.
$U(n) cap H(n)$ has similar properties to the orthogonal group so this might help.
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
answered Dec 29 '18 at 15:45
Or KedarOr Kedar
40217
40217
add a comment |
add a comment |
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$begingroup$
May I ask what is the motivation for this?
$endgroup$
– Torsten Schoeneberg
Dec 29 '18 at 19:38
1
$begingroup$
Sure. The unitary group acts transitively on Lagrangian subspaces with stabilizers orthogonal matrices. I want to create an invariant of a function to the Lagrangian grassmanian and for that to be well defined I need the eigenvalues to be “oriented “ in some way.
$endgroup$
– Elad
Dec 29 '18 at 21:51
1
$begingroup$
This invariant is some version of the Maslov index ncatlab.org/nlab/show/Maslov+index
$endgroup$
– Elad
Dec 29 '18 at 22:04