Mixing
Unitary Matrices Proof
$begingroup$
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.
I am very confused about this problem. Should I first prove that $U$ is unitary and then that $U^*$A$U$ and $U^*$B$U$ are both diagonal?
Or I should assume that $U^*$A$U$ and $U^*$B$U$ are diagonal and therefoe I should prove that $U$ is unitary??
linear-algebra matrices diagonalization
asked Jan 11 at 16:47
KaiKai
446
$endgroup$
add a comment |
$begingroup$
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.
I am very confused about this problem. Should I first prove that $U$ is unitary and then that $U^*$A$U$ and $U^*$B$U$ are both diagonal?
Or I should assume that $U^*$A$U$ and $U^*$B$U$ are diagonal and therefoe I should prove that $U$ is unitary??
linear-algebra matrices diagonalization
asked Jan 11 at 16:47
KaiKai
446
$endgroup$
add a comment |
$begingroup$
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.
I am very confused about this problem. Should I first prove that $U$ is unitary and then that $U^*$A$U$ and $U^*$B$U$ are both diagonal?
Or I should assume that $U^*$A$U$ and $U^*$B$U$ are diagonal and therefoe I should prove that $U$ is unitary??
linear-algebra matrices diagonalization
asked Jan 11 at 16:47
KaiKai
446
$endgroup$
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.
I am very confused about this problem. Should I first prove that $U$ is unitary and then that $U^*$A$U$ and $U^*$B$U$ are both diagonal?
Or I should assume that $U^*$A$U$ and $U^*$B$U$ are diagonal and therefoe I should prove that $U$ is unitary??
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
asked Jan 11 at 16:47
KaiKai
446
asked Jan 11 at 16:47
KaiKai
446
asked Jan 11 at 16:47
KaiKai
446
asked Jan 11 at 16:47
KaiKai
446
asked Jan 11 at 16:47
KaiKai
446
446
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint: Since $A:mathbb{C}^ntomathbb{C}^n$ is self-adjoint, we can find an orthonormal basis composed of eigenvectors of $A$ by spectral theorem. Thus we can decompose
$$
mathbb{C}^n = bigoplus_{i=1}^k ker(A-lambda_iI).
$$ Let $N_i = ker (A-lambda_i)$ and show that $B_i =B|_{N_i}:N_ito N_i$ is a well-defined self-adjoint operator. Find an orthonormal basis $mathcal{B}_i$ of $N_i$ consisting of eigenvectors of $B_i$ for each $i$. Finally prove that $mathcal{B}=bigcup_{i=1}^k mathcal{B}_i$ forms an orthonormal basis of $mathbb{C}^n$ consisting of eigenvectors of both $A$ and $B$. And form a desired unitary matrix $U$ using the orthonormal basis $mathcal{B}$.
answered Jan 11 at 17:50
SongSong
12.2k630
$endgroup$
1
$begingroup$
I cant understand how to use your hint for my proof. Can you give me deeper explanation?
$endgroup$
– Kai
Jan 11 at 18:10
add a comment |
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1 Answer
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$begingroup$
Hint: Since $A:mathbb{C}^ntomathbb{C}^n$ is self-adjoint, we can find an orthonormal basis composed of eigenvectors of $A$ by spectral theorem. Thus we can decompose
$$
mathbb{C}^n = bigoplus_{i=1}^k ker(A-lambda_iI).
$$ Let $N_i = ker (A-lambda_i)$ and show that $B_i =B|_{N_i}:N_ito N_i$ is a well-defined self-adjoint operator. Find an orthonormal basis $mathcal{B}_i$ of $N_i$ consisting of eigenvectors of $B_i$ for each $i$. Finally prove that $mathcal{B}=bigcup_{i=1}^k mathcal{B}_i$ forms an orthonormal basis of $mathbb{C}^n$ consisting of eigenvectors of both $A$ and $B$. And form a desired unitary matrix $U$ using the orthonormal basis $mathcal{B}$.
answered Jan 11 at 17:50
SongSong
12.2k630
$endgroup$
1
$begingroup$
I cant understand how to use your hint for my proof. Can you give me deeper explanation?
$endgroup$
– Kai
Jan 11 at 18:10
add a comment |
$begingroup$
Hint: Since $A:mathbb{C}^ntomathbb{C}^n$ is self-adjoint, we can find an orthonormal basis composed of eigenvectors of $A$ by spectral theorem. Thus we can decompose
$$
mathbb{C}^n = bigoplus_{i=1}^k ker(A-lambda_iI).
$$ Let $N_i = ker (A-lambda_i)$ and show that $B_i =B|_{N_i}:N_ito N_i$ is a well-defined self-adjoint operator. Find an orthonormal basis $mathcal{B}_i$ of $N_i$ consisting of eigenvectors of $B_i$ for each $i$. Finally prove that $mathcal{B}=bigcup_{i=1}^k mathcal{B}_i$ forms an orthonormal basis of $mathbb{C}^n$ consisting of eigenvectors of both $A$ and $B$. And form a desired unitary matrix $U$ using the orthonormal basis $mathcal{B}$.
answered Jan 11 at 17:50
SongSong
12.2k630
$endgroup$
1
$begingroup$
I cant understand how to use your hint for my proof. Can you give me deeper explanation?
$endgroup$
– Kai
Jan 11 at 18:10
add a comment |
$begingroup$
Hint: Since $A:mathbb{C}^ntomathbb{C}^n$ is self-adjoint, we can find an orthonormal basis composed of eigenvectors of $A$ by spectral theorem. Thus we can decompose
$$
mathbb{C}^n = bigoplus_{i=1}^k ker(A-lambda_iI).
$$ Let $N_i = ker (A-lambda_i)$ and show that $B_i =B|_{N_i}:N_ito N_i$ is a well-defined self-adjoint operator. Find an orthonormal basis $mathcal{B}_i$ of $N_i$ consisting of eigenvectors of $B_i$ for each $i$. Finally prove that $mathcal{B}=bigcup_{i=1}^k mathcal{B}_i$ forms an orthonormal basis of $mathbb{C}^n$ consisting of eigenvectors of both $A$ and $B$. And form a desired unitary matrix $U$ using the orthonormal basis $mathcal{B}$.
answered Jan 11 at 17:50
SongSong
12.2k630
$endgroup$
Hint: Since $A:mathbb{C}^ntomathbb{C}^n$ is self-adjoint, we can find an orthonormal basis composed of eigenvectors of $A$ by spectral theorem. Thus we can decompose
$$
mathbb{C}^n = bigoplus_{i=1}^k ker(A-lambda_iI).
$$ Let $N_i = ker (A-lambda_i)$ and show that $B_i =B|_{N_i}:N_ito N_i$ is a well-defined self-adjoint operator. Find an orthonormal basis $mathcal{B}_i$ of $N_i$ consisting of eigenvectors of $B_i$ for each $i$. Finally prove that $mathcal{B}=bigcup_{i=1}^k mathcal{B}_i$ forms an orthonormal basis of $mathbb{C}^n$ consisting of eigenvectors of both $A$ and $B$. And form a desired unitary matrix $U$ using the orthonormal basis $mathcal{B}$.
answered Jan 11 at 17:50
SongSong
12.2k630
edited Jan 11 at 21:00
answered Jan 11 at 17:50
SongSong
12.2k630
answered Jan 11 at 17:50
SongSong
12.2k630
answered Jan 11 at 17:50
SongSong
12.2k630
12.2k630
1
$begingroup$
I cant understand how to use your hint for my proof. Can you give me deeper explanation?
$endgroup$
– Kai
Jan 11 at 18:10
add a comment |
1
$begingroup$
I cant understand how to use your hint for my proof. Can you give me deeper explanation?
$endgroup$
– Kai
Jan 11 at 18:10
1
1
$begingroup$
I cant understand how to use your hint for my proof. Can you give me deeper explanation?
$endgroup$
– Kai
Jan 11 at 18:10
$begingroup$
I cant understand how to use your hint for my proof. Can you give me deeper explanation?
$endgroup$
– Kai
Jan 11 at 18:10
add a comment |
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