Diagonal Matrix Problem
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Could someone check if the solution of the problem is right?
Problem:
Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint ,such that $[A,B] :=
AB − BA = 0$
Show that there is a unitary matrix $U in mathbb{C}^{ntimes n}$ such
that $U^*$A$U$ and $U^*$$B$$U$ are both diagonal.
Solution:
Let $D$1 =$U^*$A$U$ and $D$2 =$U^*$$B$$U$ .
As $A$ and $B$ are selfadjoint follows that $D^*$=$(U^*$$A$$U)^*$= $U^*$$A$$U$=$D$1, so $D$1 is hermitian, which also means that $D$1 must be diagonal ?
linear-algebra diagonalization
asked Jan 13 at 15:07
KaiKai
446
$endgroup$
add a comment |
$begingroup$
Could someone check if the solution of the problem is right?
Problem:
Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint ,such that $[A,B] :=
AB − BA = 0$
Show that there is a unitary matrix $U in mathbb{C}^{ntimes n}$ such
that $U^*$A$U$ and $U^*$$B$$U$ are both diagonal.
Solution:
Let $D$1 =$U^*$A$U$ and $D$2 =$U^*$$B$$U$ .
As $A$ and $B$ are selfadjoint follows that $D^*$=$(U^*$$A$$U)^*$= $U^*$$A$$U$=$D$1, so $D$1 is hermitian, which also means that $D$1 must be diagonal ?
linear-algebra diagonalization
asked Jan 13 at 15:07
KaiKai
446
$endgroup$
$begingroup$
No it's wrong. you don't have the same $U$ for $A$ and $B$ in the beginning.
$endgroup$
– Yanko
Jan 13 at 15:10
add a comment |
$begingroup$
Could someone check if the solution of the problem is right?
Problem:
Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint ,such that $[A,B] :=
AB − BA = 0$
Show that there is a unitary matrix $U in mathbb{C}^{ntimes n}$ such
that $U^*$A$U$ and $U^*$$B$$U$ are both diagonal.
Solution:
Let $D$1 =$U^*$A$U$ and $D$2 =$U^*$$B$$U$ .
As $A$ and $B$ are selfadjoint follows that $D^*$=$(U^*$$A$$U)^*$= $U^*$$A$$U$=$D$1, so $D$1 is hermitian, which also means that $D$1 must be diagonal ?
linear-algebra diagonalization
asked Jan 13 at 15:07
KaiKai
446
$endgroup$
Could someone check if the solution of the problem is right?
Problem:
Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint ,such that $[A,B] :=
AB − BA = 0$
Show that there is a unitary matrix $U in mathbb{C}^{ntimes n}$ such
that $U^*$A$U$ and $U^*$$B$$U$ are both diagonal.
Solution:
Let $D$1 =$U^*$A$U$ and $D$2 =$U^*$$B$$U$ .
As $A$ and $B$ are selfadjoint follows that $D^*$=$(U^*$$A$$U)^*$= $U^*$$A$$U$=$D$1, so $D$1 is hermitian, which also means that $D$1 must be diagonal ?
linear-algebra diagonalization
linear-algebra diagonalization
asked Jan 13 at 15:07
KaiKai
446
asked Jan 13 at 15:07
KaiKai
446
asked Jan 13 at 15:07
KaiKai
446
asked Jan 13 at 15:07
KaiKai
446
asked Jan 13 at 15:07
KaiKai
446
446
$begingroup$
No it's wrong. you don't have the same $U$ for $A$ and $B$ in the beginning.
$endgroup$
– Yanko
Jan 13 at 15:10
add a comment |
$begingroup$
No it's wrong. you don't have the same $U$ for $A$ and $B$ in the beginning.
$endgroup$
– Yanko
Jan 13 at 15:10
$begingroup$
No it's wrong. you don't have the same $U$ for $A$ and $B$ in the beginning.
$endgroup$
– Yanko
Jan 13 at 15:10
$begingroup$
No it's wrong. you don't have the same $U$ for $A$ and $B$ in the beginning.
$endgroup$
– Yanko
Jan 13 at 15:10
add a comment |
1 Answer
1
active
oldest
votes
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No, it is not right. You started your solution by saying “Let $D_1=U^*AU$”, without saying what $U$ is.
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
In the problem is written that U is a unitray matrix.
$endgroup$
– Kai
Jan 13 at 15:12
1
$begingroup$
@Kai The problem asks you to find such unitary $U$.
$endgroup$
– xbh
Jan 13 at 15:16
$begingroup$
@ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix?
$endgroup$
– Kai
Jan 13 at 15:25
$begingroup$
@Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal.
$endgroup$
– José Carlos Santos
Jan 13 at 15:32
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it is not right. You started your solution by saying “Let $D_1=U^*AU$”, without saying what $U$ is.
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
In the problem is written that U is a unitray matrix.
$endgroup$
– Kai
Jan 13 at 15:12
1
$begingroup$
@Kai The problem asks you to find such unitary $U$.
$endgroup$
– xbh
Jan 13 at 15:16
$begingroup$
@ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix?
$endgroup$
– Kai
Jan 13 at 15:25
$begingroup$
@Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal.
$endgroup$
– José Carlos Santos
Jan 13 at 15:32
add a comment |
$begingroup$
No, it is not right. You started your solution by saying “Let $D_1=U^*AU$”, without saying what $U$ is.
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
In the problem is written that U is a unitray matrix.
$endgroup$
– Kai
Jan 13 at 15:12
1
$begingroup$
@Kai The problem asks you to find such unitary $U$.
$endgroup$
– xbh
Jan 13 at 15:16
$begingroup$
@ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix?
$endgroup$
– Kai
Jan 13 at 15:25
$begingroup$
@Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal.
$endgroup$
– José Carlos Santos
Jan 13 at 15:32
add a comment |
$begingroup$
No, it is not right. You started your solution by saying “Let $D_1=U^*AU$”, without saying what $U$ is.
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
No, it is not right. You started your solution by saying “Let $D_1=U^*AU$”, without saying what $U$ is.
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 13 at 15:10
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
$begingroup$
In the problem is written that U is a unitray matrix.
$endgroup$
– Kai
Jan 13 at 15:12
1
$begingroup$
@Kai The problem asks you to find such unitary $U$.
$endgroup$
– xbh
Jan 13 at 15:16
$begingroup$
@ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix?
$endgroup$
– Kai
Jan 13 at 15:25
$begingroup$
@Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal.
$endgroup$
– José Carlos Santos
Jan 13 at 15:32
add a comment |
$begingroup$
In the problem is written that U is a unitray matrix.
$endgroup$
– Kai
Jan 13 at 15:12
1
$begingroup$
@Kai The problem asks you to find such unitary $U$.
$endgroup$
– xbh
Jan 13 at 15:16
$begingroup$
@ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix?
$endgroup$
– Kai
Jan 13 at 15:25
$begingroup$
@Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal.
$endgroup$
– José Carlos Santos
Jan 13 at 15:32
$begingroup$
In the problem is written that U is a unitray matrix.
$endgroup$
– Kai
Jan 13 at 15:12
$begingroup$
In the problem is written that U is a unitray matrix.
$endgroup$
– Kai
Jan 13 at 15:12
1
1
$begingroup$
@Kai The problem asks you to find such unitary $U$.
$endgroup$
– xbh
Jan 13 at 15:16
$begingroup$
@Kai The problem asks you to find such unitary $U$.
$endgroup$
– xbh
Jan 13 at 15:16
$begingroup$
@ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix?
$endgroup$
– Kai
Jan 13 at 15:25
$begingroup$
@ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix?
$endgroup$
– Kai
Jan 13 at 15:25
$begingroup$
@Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal.
$endgroup$
– José Carlos Santos
Jan 13 at 15:32
$begingroup$
@Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal.
$endgroup$
– José Carlos Santos
Jan 13 at 15:32
add a comment |
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$begingroup$
No it's wrong. you don't have the same $U$ for $A$ and $B$ in the beginning.
$endgroup$
– Yanko
Jan 13 at 15:10