Mixing
if $AA^*=BB^*$ what are the relations between A and B [closed]
$begingroup$
I'm wondering if we have two linear operators $A, B in ell(V)$. and we know that $AA^*=BB^*$. then what informations can this give to us about relationships between $A$ and $B$?
I think they have some strong relations.
linear-algebra operator-theory adjoint-operators
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
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closed as off-topic by Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj Jan 13 at 11:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm wondering if we have two linear operators $A, B in ell(V)$. and we know that $AA^*=BB^*$. then what informations can this give to us about relationships between $A$ and $B$?
I think they have some strong relations.
linear-algebra operator-theory adjoint-operators
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
closed as off-topic by Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj Jan 13 at 11:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
In view of polar decomposition, it tells that $A=PU$ and $B=PU'$ for $P = sqrt{AA^*} = sqrt{BB^*}$ and $U, U'$ some unitary operators. So, there exists a unitary operators $V$ such that $B=AV$.
$endgroup$
– Sangchul Lee
Jan 13 at 6:00
$begingroup$
@SangchulLee great! I think I've got my answer. thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 6:06
add a comment |
$begingroup$
I'm wondering if we have two linear operators $A, B in ell(V)$. and we know that $AA^*=BB^*$. then what informations can this give to us about relationships between $A$ and $B$?
I think they have some strong relations.
linear-algebra operator-theory adjoint-operators
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
I'm wondering if we have two linear operators $A, B in ell(V)$. and we know that $AA^*=BB^*$. then what informations can this give to us about relationships between $A$ and $B$?
I think they have some strong relations.
linear-algebra operator-theory adjoint-operators
linear-algebra operator-theory adjoint-operators
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
edited Jan 13 at 5:51
Peyman mohseni kiasari
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 13 at 5:28
Peyman mohseni kiasariPeyman mohseni kiasari
1089
1089
closed as off-topic by Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj Jan 13 at 11:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj Jan 13 at 11:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Leucippus, Shailesh, Cesareo, Nikunj
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
In view of polar decomposition, it tells that $A=PU$ and $B=PU'$ for $P = sqrt{AA^*} = sqrt{BB^*}$ and $U, U'$ some unitary operators. So, there exists a unitary operators $V$ such that $B=AV$.
$endgroup$
– Sangchul Lee
Jan 13 at 6:00
$begingroup$
@SangchulLee great! I think I've got my answer. thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 6:06
add a comment |
1
$begingroup$
In view of polar decomposition, it tells that $A=PU$ and $B=PU'$ for $P = sqrt{AA^*} = sqrt{BB^*}$ and $U, U'$ some unitary operators. So, there exists a unitary operators $V$ such that $B=AV$.
$endgroup$
– Sangchul Lee
Jan 13 at 6:00
$begingroup$
@SangchulLee great! I think I've got my answer. thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 6:06
1
1
$begingroup$
In view of polar decomposition, it tells that $A=PU$ and $B=PU'$ for $P = sqrt{AA^*} = sqrt{BB^*}$ and $U, U'$ some unitary operators. So, there exists a unitary operators $V$ such that $B=AV$.
$endgroup$
– Sangchul Lee
Jan 13 at 6:00
$begingroup$
In view of polar decomposition, it tells that $A=PU$ and $B=PU'$ for $P = sqrt{AA^*} = sqrt{BB^*}$ and $U, U'$ some unitary operators. So, there exists a unitary operators $V$ such that $B=AV$.
$endgroup$
– Sangchul Lee
Jan 13 at 6:00
$begingroup$
@SangchulLee great! I think I've got my answer. thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 6:06
$begingroup$
@SangchulLee great! I think I've got my answer. thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 6:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
in finite dimension by polar decomposition, $A = sqrt{AA^*}U_a, B = sqrt{BB^*}U_b$, where the $U_a$ and $U_b$ are unitary operators. so we heve:
$sqrt{BB^*} = sqrt{AA^*} = AU^*_a = BU^*_b Rightarrow A = BU^*_bU_a$
$U^*_bU_a(U^*_bU_a)^* = U^*_bU_aU^*_aU_b = I Rightarrow U = U^*_bU_a$ is unitary
$Rightarrow A = BU$ where $U$ is an unitary operator.
notice that if $A = BU$ then $A^* = U^*B^*$ so if we had $A^*A = B^*B$ then $A = U'B$ where $U'$ is an unitary operator.
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
$begingroup$
This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry.
$endgroup$
– Martin Argerami
Jan 13 at 23:53
$begingroup$
@MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
in finite dimension by polar decomposition, $A = sqrt{AA^*}U_a, B = sqrt{BB^*}U_b$, where the $U_a$ and $U_b$ are unitary operators. so we heve:
$sqrt{BB^*} = sqrt{AA^*} = AU^*_a = BU^*_b Rightarrow A = BU^*_bU_a$
$U^*_bU_a(U^*_bU_a)^* = U^*_bU_aU^*_aU_b = I Rightarrow U = U^*_bU_a$ is unitary
$Rightarrow A = BU$ where $U$ is an unitary operator.
notice that if $A = BU$ then $A^* = U^*B^*$ so if we had $A^*A = B^*B$ then $A = U'B$ where $U'$ is an unitary operator.
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
$begingroup$
This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry.
$endgroup$
– Martin Argerami
Jan 13 at 23:53
$begingroup$
@MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:56
add a comment |
$begingroup$
in finite dimension by polar decomposition, $A = sqrt{AA^*}U_a, B = sqrt{BB^*}U_b$, where the $U_a$ and $U_b$ are unitary operators. so we heve:
$sqrt{BB^*} = sqrt{AA^*} = AU^*_a = BU^*_b Rightarrow A = BU^*_bU_a$
$U^*_bU_a(U^*_bU_a)^* = U^*_bU_aU^*_aU_b = I Rightarrow U = U^*_bU_a$ is unitary
$Rightarrow A = BU$ where $U$ is an unitary operator.
notice that if $A = BU$ then $A^* = U^*B^*$ so if we had $A^*A = B^*B$ then $A = U'B$ where $U'$ is an unitary operator.
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
$begingroup$
This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry.
$endgroup$
– Martin Argerami
Jan 13 at 23:53
$begingroup$
@MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:56
add a comment |
$begingroup$
in finite dimension by polar decomposition, $A = sqrt{AA^*}U_a, B = sqrt{BB^*}U_b$, where the $U_a$ and $U_b$ are unitary operators. so we heve:
$sqrt{BB^*} = sqrt{AA^*} = AU^*_a = BU^*_b Rightarrow A = BU^*_bU_a$
$U^*_bU_a(U^*_bU_a)^* = U^*_bU_aU^*_aU_b = I Rightarrow U = U^*_bU_a$ is unitary
$Rightarrow A = BU$ where $U$ is an unitary operator.
notice that if $A = BU$ then $A^* = U^*B^*$ so if we had $A^*A = B^*B$ then $A = U'B$ where $U'$ is an unitary operator.
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
in finite dimension by polar decomposition, $A = sqrt{AA^*}U_a, B = sqrt{BB^*}U_b$, where the $U_a$ and $U_b$ are unitary operators. so we heve:
$sqrt{BB^*} = sqrt{AA^*} = AU^*_a = BU^*_b Rightarrow A = BU^*_bU_a$
$U^*_bU_a(U^*_bU_a)^* = U^*_bU_aU^*_aU_b = I Rightarrow U = U^*_bU_a$ is unitary
$Rightarrow A = BU$ where $U$ is an unitary operator.
notice that if $A = BU$ then $A^* = U^*B^*$ so if we had $A^*A = B^*B$ then $A = U'B$ where $U'$ is an unitary operator.
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
edited Jan 13 at 23:54
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
answered Jan 13 at 6:24
Peyman mohseni kiasariPeyman mohseni kiasari
1089
1089
$begingroup$
This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry.
$endgroup$
– Martin Argerami
Jan 13 at 23:53
$begingroup$
@MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:56
add a comment |
$begingroup$
This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry.
$endgroup$
– Martin Argerami
Jan 13 at 23:53
$begingroup$
@MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:56
$begingroup$
This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry.
$endgroup$
– Martin Argerami
Jan 13 at 23:53
$begingroup$
This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry.
$endgroup$
– Martin Argerami
Jan 13 at 23:53
$begingroup$
@MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:56
$begingroup$
@MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:56
add a comment |
1
$begingroup$
In view of polar decomposition, it tells that $A=PU$ and $B=PU'$ for $P = sqrt{AA^*} = sqrt{BB^*}$ and $U, U'$ some unitary operators. So, there exists a unitary operators $V$ such that $B=AV$.
$endgroup$
– Sangchul Lee
Jan 13 at 6:00
$begingroup$
@SangchulLee great! I think I've got my answer. thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 6:06