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Conditions to obtain a real logarithm of a unitary unimodular complex matrix?
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The problem statement is the following:
$$U=expiV$$
where $U$ is a unitary unimodular matrix of the following form:
$$U=beginbmatrixu_1+iu_2&u_3+iu_4\-u_3+iu_4&u_1-iu_2endbmatrixinmathbbC^2times2$$
with
$$u_1^2+u_2^2+u_3^2+u_4^2=1, u_jinmathbbR forall jin1,...,4$$
and where $VinmathbbR^2times2$, and $i$ is the imaginary unit.
I am looking for solutions $VinmathbbR^2times2$ of this problem. What conditions, in general, must be fulfilled for the logarithm of $U$ to be a real matrix i.e.:
$$-ilogU=VinmathbbR^2times2$$
real-analysis linear-algebra functional-analysis exponential-function matrix-exponential
asked Jul 17 at 8:47
john melon
472312
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up vote
0
down vote
favorite
The problem statement is the following:
$$U=expiV$$
where $U$ is a unitary unimodular matrix of the following form:
$$U=beginbmatrixu_1+iu_2&u_3+iu_4\-u_3+iu_4&u_1-iu_2endbmatrixinmathbbC^2times2$$
with
$$u_1^2+u_2^2+u_3^2+u_4^2=1, u_jinmathbbR forall jin1,...,4$$
and where $VinmathbbR^2times2$, and $i$ is the imaginary unit.
I am looking for solutions $VinmathbbR^2times2$ of this problem. What conditions, in general, must be fulfilled for the logarithm of $U$ to be a real matrix i.e.:
$$-ilogU=VinmathbbR^2times2$$
real-analysis linear-algebra functional-analysis exponential-function matrix-exponential
asked Jul 17 at 8:47
john melon
472312
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The problem statement is the following:
$$U=expiV$$
where $U$ is a unitary unimodular matrix of the following form:
$$U=beginbmatrixu_1+iu_2&u_3+iu_4\-u_3+iu_4&u_1-iu_2endbmatrixinmathbbC^2times2$$
with
$$u_1^2+u_2^2+u_3^2+u_4^2=1, u_jinmathbbR forall jin1,...,4$$
and where $VinmathbbR^2times2$, and $i$ is the imaginary unit.
I am looking for solutions $VinmathbbR^2times2$ of this problem. What conditions, in general, must be fulfilled for the logarithm of $U$ to be a real matrix i.e.:
$$-ilogU=VinmathbbR^2times2$$
real-analysis linear-algebra functional-analysis exponential-function matrix-exponential
asked Jul 17 at 8:47
john melon
472312
The problem statement is the following:
$$U=expiV$$
where $U$ is a unitary unimodular matrix of the following form:
$$U=beginbmatrixu_1+iu_2&u_3+iu_4\-u_3+iu_4&u_1-iu_2endbmatrixinmathbbC^2times2$$
with
$$u_1^2+u_2^2+u_3^2+u_4^2=1, u_jinmathbbR forall jin1,...,4$$
and where $VinmathbbR^2times2$, and $i$ is the imaginary unit.
I am looking for solutions $VinmathbbR^2times2$ of this problem. What conditions, in general, must be fulfilled for the logarithm of $U$ to be a real matrix i.e.:
$$-ilogU=VinmathbbR^2times2$$
real-analysis linear-algebra functional-analysis exponential-function matrix-exponential
asked Jul 17 at 8:47
john melon
472312
asked Jul 17 at 8:47
john melon
472312
asked Jul 17 at 8:47
john melon
472312
asked Jul 17 at 8:47
john melon
472312
472312
add a comment |Â
add a comment |Â
1 Answer
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1
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The conditions are $det(U)=1,U^*=U^-1$ and $U=e^iV$ where $V$ is real.
Then $overlineU=e^-iV=U^-1=U^*$, that is $U=U^T$. Then, necessarily $u_3=0$.
Conversely, assume that $u_3=0$. We may use the principal log when $U$ has no $leq 0$ eigenvalues, that is here when $-1notin spectrum(U)$, that is when $u_1not= -1$. Then $-ilog(U)$ is a real solution in the form $beginpmatrixr&p\p&-rendpmatrix$ (symmetric as $U$ and with zero trace).
If $u_1=-1$, then the other $u_i$ are $0$ and $U=-I_2$. We cannot no more use the principal log; yet a real solution is $V=pi I_2$.
answered Jul 18 at 10:34
loup blanc
20.3k21549
Thank you for your response. Why do the other $u_i$ have to be zero in the case when $u_1=-1$?
– john melon
Jul 18 at 13:01
because $sum_i u_i^2=1$.
– loup blanc
Jul 18 at 13:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The conditions are $det(U)=1,U^*=U^-1$ and $U=e^iV$ where $V$ is real.
Then $overlineU=e^-iV=U^-1=U^*$, that is $U=U^T$. Then, necessarily $u_3=0$.
Conversely, assume that $u_3=0$. We may use the principal log when $U$ has no $leq 0$ eigenvalues, that is here when $-1notin spectrum(U)$, that is when $u_1not= -1$. Then $-ilog(U)$ is a real solution in the form $beginpmatrixr&p\p&-rendpmatrix$ (symmetric as $U$ and with zero trace).
If $u_1=-1$, then the other $u_i$ are $0$ and $U=-I_2$. We cannot no more use the principal log; yet a real solution is $V=pi I_2$.
answered Jul 18 at 10:34
loup blanc
20.3k21549
Thank you for your response. Why do the other $u_i$ have to be zero in the case when $u_1=-1$?
– john melon
Jul 18 at 13:01
because $sum_i u_i^2=1$.
– loup blanc
Jul 18 at 13:12
add a comment |Â
up vote
1
down vote
accepted
The conditions are $det(U)=1,U^*=U^-1$ and $U=e^iV$ where $V$ is real.
Then $overlineU=e^-iV=U^-1=U^*$, that is $U=U^T$. Then, necessarily $u_3=0$.
Conversely, assume that $u_3=0$. We may use the principal log when $U$ has no $leq 0$ eigenvalues, that is here when $-1notin spectrum(U)$, that is when $u_1not= -1$. Then $-ilog(U)$ is a real solution in the form $beginpmatrixr&p\p&-rendpmatrix$ (symmetric as $U$ and with zero trace).
If $u_1=-1$, then the other $u_i$ are $0$ and $U=-I_2$. We cannot no more use the principal log; yet a real solution is $V=pi I_2$.
answered Jul 18 at 10:34
loup blanc
20.3k21549
Thank you for your response. Why do the other $u_i$ have to be zero in the case when $u_1=-1$?
– john melon
Jul 18 at 13:01
because $sum_i u_i^2=1$.
– loup blanc
Jul 18 at 13:12
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The conditions are $det(U)=1,U^*=U^-1$ and $U=e^iV$ where $V$ is real.
Then $overlineU=e^-iV=U^-1=U^*$, that is $U=U^T$. Then, necessarily $u_3=0$.
Conversely, assume that $u_3=0$. We may use the principal log when $U$ has no $leq 0$ eigenvalues, that is here when $-1notin spectrum(U)$, that is when $u_1not= -1$. Then $-ilog(U)$ is a real solution in the form $beginpmatrixr&p\p&-rendpmatrix$ (symmetric as $U$ and with zero trace).
If $u_1=-1$, then the other $u_i$ are $0$ and $U=-I_2$. We cannot no more use the principal log; yet a real solution is $V=pi I_2$.
answered Jul 18 at 10:34
loup blanc
20.3k21549
The conditions are $det(U)=1,U^*=U^-1$ and $U=e^iV$ where $V$ is real.
Then $overlineU=e^-iV=U^-1=U^*$, that is $U=U^T$. Then, necessarily $u_3=0$.
Conversely, assume that $u_3=0$. We may use the principal log when $U$ has no $leq 0$ eigenvalues, that is here when $-1notin spectrum(U)$, that is when $u_1not= -1$. Then $-ilog(U)$ is a real solution in the form $beginpmatrixr&p\p&-rendpmatrix$ (symmetric as $U$ and with zero trace).
If $u_1=-1$, then the other $u_i$ are $0$ and $U=-I_2$. We cannot no more use the principal log; yet a real solution is $V=pi I_2$.
answered Jul 18 at 10:34
loup blanc
20.3k21549
answered Jul 18 at 10:34
loup blanc
20.3k21549
answered Jul 18 at 10:34
loup blanc
20.3k21549
answered Jul 18 at 10:34
loup blanc
20.3k21549
20.3k21549
Thank you for your response. Why do the other $u_i$ have to be zero in the case when $u_1=-1$?
– john melon
Jul 18 at 13:01
because $sum_i u_i^2=1$.
– loup blanc
Jul 18 at 13:12
add a comment |Â
Thank you for your response. Why do the other $u_i$ have to be zero in the case when $u_1=-1$?
– john melon
Jul 18 at 13:01
because $sum_i u_i^2=1$.
– loup blanc
Jul 18 at 13:12
Thank you for your response. Why do the other $u_i$ have to be zero in the case when $u_1=-1$?
– john melon
Jul 18 at 13:01
Thank you for your response. Why do the other $u_i$ have to be zero in the case when $u_1=-1$?
– john melon
Jul 18 at 13:01
because $sum_i u_i^2=1$.
– loup blanc
Jul 18 at 13:12
because $sum_i u_i^2=1$.
– loup blanc
Jul 18 at 13:12
add a comment |Â
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