When will $AB=BA$?
$begingroup$
Given two square matrices $A,B$ with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite simple question, but I can find little information about it. Thanks.
matrices
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
$endgroup$
add a comment |
$begingroup$
Given two square matrices $A,B$ with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite simple question, but I can find little information about it. Thanks.
matrices
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
$endgroup$
5
$begingroup$
en.wikipedia.org/wiki/Commuting_matrices
$endgroup$
– B0rk4
Aug 29 '13 at 7:30
4
$begingroup$
Since everybody (except Hauke) is just listing their favorite sufficient conditions let me add mine: If there exists a polynomial $Pin R[X]$ ($R$ a commutative ring containing the entries of $A$ and $B$) such that $B=P(A)$, then we have $AB=BA$. Furthermore, in the case that $R$ is (contained in) an algebraically closed field and the eigenvalues of $A$ are distinct, then this sufficient criterion is also necessary. For more read the Wikiarticle linked to by Hauke.
$endgroup$
– Jyrki Lahtonen
Aug 29 '13 at 8:17
add a comment |
$begingroup$
Given two square matrices $A,B$ with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite simple question, but I can find little information about it. Thanks.
matrices
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
$endgroup$
Given two square matrices $A,B$ with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite simple question, but I can find little information about it. Thanks.
matrices
matrices
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
asked Aug 29 '13 at 7:11
A. ChuA. Chu
7,05093485
7,05093485
5
$begingroup$
en.wikipedia.org/wiki/Commuting_matrices
$endgroup$
– B0rk4
Aug 29 '13 at 7:30
4
$begingroup$
Since everybody (except Hauke) is just listing their favorite sufficient conditions let me add mine: If there exists a polynomial $Pin R[X]$ ($R$ a commutative ring containing the entries of $A$ and $B$) such that $B=P(A)$, then we have $AB=BA$. Furthermore, in the case that $R$ is (contained in) an algebraically closed field and the eigenvalues of $A$ are distinct, then this sufficient criterion is also necessary. For more read the Wikiarticle linked to by Hauke.
$endgroup$
– Jyrki Lahtonen
Aug 29 '13 at 8:17
add a comment |
5
$begingroup$
en.wikipedia.org/wiki/Commuting_matrices
$endgroup$
– B0rk4
Aug 29 '13 at 7:30
4
$begingroup$
Since everybody (except Hauke) is just listing their favorite sufficient conditions let me add mine: If there exists a polynomial $Pin R[X]$ ($R$ a commutative ring containing the entries of $A$ and $B$) such that $B=P(A)$, then we have $AB=BA$. Furthermore, in the case that $R$ is (contained in) an algebraically closed field and the eigenvalues of $A$ are distinct, then this sufficient criterion is also necessary. For more read the Wikiarticle linked to by Hauke.
$endgroup$
– Jyrki Lahtonen
Aug 29 '13 at 8:17
5
5
$begingroup$
en.wikipedia.org/wiki/Commuting_matrices
$endgroup$
– B0rk4
Aug 29 '13 at 7:30
$begingroup$
en.wikipedia.org/wiki/Commuting_matrices
$endgroup$
– B0rk4
Aug 29 '13 at 7:30
4
4
$begingroup$
Since everybody (except Hauke) is just listing their favorite sufficient conditions let me add mine: If there exists a polynomial $Pin R[X]$ ($R$ a commutative ring containing the entries of $A$ and $B$) such that $B=P(A)$, then we have $AB=BA$. Furthermore, in the case that $R$ is (contained in) an algebraically closed field and the eigenvalues of $A$ are distinct, then this sufficient criterion is also necessary. For more read the Wikiarticle linked to by Hauke.
$endgroup$
– Jyrki Lahtonen
Aug 29 '13 at 8:17
$begingroup$
Since everybody (except Hauke) is just listing their favorite sufficient conditions let me add mine: If there exists a polynomial $Pin R[X]$ ($R$ a commutative ring containing the entries of $A$ and $B$) such that $B=P(A)$, then we have $AB=BA$. Furthermore, in the case that $R$ is (contained in) an algebraically closed field and the eigenvalues of $A$ are distinct, then this sufficient criterion is also necessary. For more read the Wikiarticle linked to by Hauke.
$endgroup$
– Jyrki Lahtonen
Aug 29 '13 at 8:17
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If $A,B$ are diagonalizable, they commute if and only if they are simultaneously diagonalizable. For a proof, see here. This, of course, means that they have a common set of eigenvectors.
If $A,B$ are normal (i.e., unitarily diagonalizable), they commute if and only if they are simultaneously unitarily diagonalizable. A proof can be done by using the Schur decomposition of a commuting family. This, of course, means that they have a common set of orthonormal eigenvectors.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
$endgroup$
add a comment |
$begingroup$
This is too long for a comment, so I posted it as an answer.
I think it really depends on what $A$ or $B$ is. For example, if $A=cI$ where $I$ is the identity matrix, then $AB=BA$ for all matrices $B$. In fact, the converse is true:
If $A$ is an $ntimes n$ matrix such that $AB=BA$ for all $ntimes n$ matrices $B$, then $A=c I$ for some constant $c$.
Therefore, if $A$ is not in the form of $c I$, there must be some matrix $B$ such that $ABneq BA$.
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
$endgroup$
add a comment |
$begingroup$
Here are some different cases I can think of:
- $A=B$.
- Either $A=cI$ or $B=cI$, as already stated by Paul.
- $A$ and $B$ are both diagonal matrices.
- There exists an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal.
edited Nov 17 '17 at 5:03
Community♦
1
answered Aug 29 '13 at 7:31
RyanRyan
596628
$endgroup$
add a comment |
$begingroup$
There is actually a sufficient and necessary condition for $M_n(mathbb{C})$:
Let $J$ be the Jordan canonical form of a complex matrix $A$, i.e.,
$$
A=PJP^{-1}=Pmathrm{diag}(J_1,cdots,J_s)P^{-1}
$$
where
$$
J_i=lambda_iI+N_i=left(begin{matrix}
lambda_i & & &\
1 & ddots & & \
& ddots & ddots & \
& & 1 & lambda_i
end{matrix}right).
$$
Then the matrices commutable with $A$ have the form of
$$
B=PB_1P^{-1}=P(B_{ij})P^{-1}
$$
where $B_1=(B_{ij})$ has the same blocking as $J$, and
$$
B_{ij}=begin{cases}
0 &mbox{if } lambda_inelambda_j\
mbox{a }unicode{x201C}mbox{lower-triangle-layered matrix''} & mbox{if } lambda_i=lambda_j
end{cases}
$$
The reference is my textbook of linear algebra, though not in English. Here is an example in the book:
The main part in the proof is to compare the corresponding entries in both sides of the equation $N_iB_{ij}=B_{ij}N_j$ when $lambda_i=lambda_j$.
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
$endgroup$
add a comment |
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4 Answers
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active
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votes
4 Answers
4
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If $A,B$ are diagonalizable, they commute if and only if they are simultaneously diagonalizable. For a proof, see here. This, of course, means that they have a common set of eigenvectors.
If $A,B$ are normal (i.e., unitarily diagonalizable), they commute if and only if they are simultaneously unitarily diagonalizable. A proof can be done by using the Schur decomposition of a commuting family. This, of course, means that they have a common set of orthonormal eigenvectors.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
$endgroup$
add a comment |
$begingroup$
If $A,B$ are diagonalizable, they commute if and only if they are simultaneously diagonalizable. For a proof, see here. This, of course, means that they have a common set of eigenvectors.
If $A,B$ are normal (i.e., unitarily diagonalizable), they commute if and only if they are simultaneously unitarily diagonalizable. A proof can be done by using the Schur decomposition of a commuting family. This, of course, means that they have a common set of orthonormal eigenvectors.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
$endgroup$
add a comment |
$begingroup$
If $A,B$ are diagonalizable, they commute if and only if they are simultaneously diagonalizable. For a proof, see here. This, of course, means that they have a common set of eigenvectors.
If $A,B$ are normal (i.e., unitarily diagonalizable), they commute if and only if they are simultaneously unitarily diagonalizable. A proof can be done by using the Schur decomposition of a commuting family. This, of course, means that they have a common set of orthonormal eigenvectors.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
$endgroup$
If $A,B$ are diagonalizable, they commute if and only if they are simultaneously diagonalizable. For a proof, see here. This, of course, means that they have a common set of eigenvectors.
If $A,B$ are normal (i.e., unitarily diagonalizable), they commute if and only if they are simultaneously unitarily diagonalizable. A proof can be done by using the Schur decomposition of a commuting family. This, of course, means that they have a common set of orthonormal eigenvectors.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
answered Aug 29 '13 at 7:45
Vedran ŠegoVedran Šego
9,65112047
9,65112047
add a comment |
add a comment |
$begingroup$
This is too long for a comment, so I posted it as an answer.
I think it really depends on what $A$ or $B$ is. For example, if $A=cI$ where $I$ is the identity matrix, then $AB=BA$ for all matrices $B$. In fact, the converse is true:
If $A$ is an $ntimes n$ matrix such that $AB=BA$ for all $ntimes n$ matrices $B$, then $A=c I$ for some constant $c$.
Therefore, if $A$ is not in the form of $c I$, there must be some matrix $B$ such that $ABneq BA$.
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
$endgroup$
add a comment |
$begingroup$
This is too long for a comment, so I posted it as an answer.
I think it really depends on what $A$ or $B$ is. For example, if $A=cI$ where $I$ is the identity matrix, then $AB=BA$ for all matrices $B$. In fact, the converse is true:
If $A$ is an $ntimes n$ matrix such that $AB=BA$ for all $ntimes n$ matrices $B$, then $A=c I$ for some constant $c$.
Therefore, if $A$ is not in the form of $c I$, there must be some matrix $B$ such that $ABneq BA$.
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
$endgroup$
add a comment |
$begingroup$
This is too long for a comment, so I posted it as an answer.
I think it really depends on what $A$ or $B$ is. For example, if $A=cI$ where $I$ is the identity matrix, then $AB=BA$ for all matrices $B$. In fact, the converse is true:
If $A$ is an $ntimes n$ matrix such that $AB=BA$ for all $ntimes n$ matrices $B$, then $A=c I$ for some constant $c$.
Therefore, if $A$ is not in the form of $c I$, there must be some matrix $B$ such that $ABneq BA$.
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
$endgroup$
This is too long for a comment, so I posted it as an answer.
I think it really depends on what $A$ or $B$ is. For example, if $A=cI$ where $I$ is the identity matrix, then $AB=BA$ for all matrices $B$. In fact, the converse is true:
If $A$ is an $ntimes n$ matrix such that $AB=BA$ for all $ntimes n$ matrices $B$, then $A=c I$ for some constant $c$.
Therefore, if $A$ is not in the form of $c I$, there must be some matrix $B$ such that $ABneq BA$.
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
answered Aug 29 '13 at 7:17
PaulPaul
16.1k33767
16.1k33767
add a comment |
add a comment |
$begingroup$
Here are some different cases I can think of:
- $A=B$.
- Either $A=cI$ or $B=cI$, as already stated by Paul.
- $A$ and $B$ are both diagonal matrices.
- There exists an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal.
edited Nov 17 '17 at 5:03
Community♦
1
answered Aug 29 '13 at 7:31
RyanRyan
596628
$endgroup$
add a comment |
$begingroup$
Here are some different cases I can think of:
- $A=B$.
- Either $A=cI$ or $B=cI$, as already stated by Paul.
- $A$ and $B$ are both diagonal matrices.
- There exists an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal.
edited Nov 17 '17 at 5:03
Community♦
1
answered Aug 29 '13 at 7:31
RyanRyan
596628
$endgroup$
add a comment |
$begingroup$
Here are some different cases I can think of:
- $A=B$.
- Either $A=cI$ or $B=cI$, as already stated by Paul.
- $A$ and $B$ are both diagonal matrices.
- There exists an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal.
edited Nov 17 '17 at 5:03
Community♦
1
answered Aug 29 '13 at 7:31
RyanRyan
596628
$endgroup$
Here are some different cases I can think of:
- $A=B$.
- Either $A=cI$ or $B=cI$, as already stated by Paul.
- $A$ and $B$ are both diagonal matrices.
- There exists an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal.
edited Nov 17 '17 at 5:03
Community♦
1
answered Aug 29 '13 at 7:31
RyanRyan
596628
edited Nov 17 '17 at 5:03
Community♦
1
edited Nov 17 '17 at 5:03
Community♦
1
edited Nov 17 '17 at 5:03
Community♦
1
1
answered Aug 29 '13 at 7:31
RyanRyan
596628
answered Aug 29 '13 at 7:31
RyanRyan
596628
answered Aug 29 '13 at 7:31
RyanRyan
596628
596628
add a comment |
add a comment |
$begingroup$
There is actually a sufficient and necessary condition for $M_n(mathbb{C})$:
Let $J$ be the Jordan canonical form of a complex matrix $A$, i.e.,
$$
A=PJP^{-1}=Pmathrm{diag}(J_1,cdots,J_s)P^{-1}
$$
where
$$
J_i=lambda_iI+N_i=left(begin{matrix}
lambda_i & & &\
1 & ddots & & \
& ddots & ddots & \
& & 1 & lambda_i
end{matrix}right).
$$
Then the matrices commutable with $A$ have the form of
$$
B=PB_1P^{-1}=P(B_{ij})P^{-1}
$$
where $B_1=(B_{ij})$ has the same blocking as $J$, and
$$
B_{ij}=begin{cases}
0 &mbox{if } lambda_inelambda_j\
mbox{a }unicode{x201C}mbox{lower-triangle-layered matrix''} & mbox{if } lambda_i=lambda_j
end{cases}
$$
The reference is my textbook of linear algebra, though not in English. Here is an example in the book:
The main part in the proof is to compare the corresponding entries in both sides of the equation $N_iB_{ij}=B_{ij}N_j$ when $lambda_i=lambda_j$.
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
$endgroup$
add a comment |
$begingroup$
There is actually a sufficient and necessary condition for $M_n(mathbb{C})$:
Let $J$ be the Jordan canonical form of a complex matrix $A$, i.e.,
$$
A=PJP^{-1}=Pmathrm{diag}(J_1,cdots,J_s)P^{-1}
$$
where
$$
J_i=lambda_iI+N_i=left(begin{matrix}
lambda_i & & &\
1 & ddots & & \
& ddots & ddots & \
& & 1 & lambda_i
end{matrix}right).
$$
Then the matrices commutable with $A$ have the form of
$$
B=PB_1P^{-1}=P(B_{ij})P^{-1}
$$
where $B_1=(B_{ij})$ has the same blocking as $J$, and
$$
B_{ij}=begin{cases}
0 &mbox{if } lambda_inelambda_j\
mbox{a }unicode{x201C}mbox{lower-triangle-layered matrix''} & mbox{if } lambda_i=lambda_j
end{cases}
$$
The reference is my textbook of linear algebra, though not in English. Here is an example in the book:
The main part in the proof is to compare the corresponding entries in both sides of the equation $N_iB_{ij}=B_{ij}N_j$ when $lambda_i=lambda_j$.
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
$endgroup$
add a comment |
$begingroup$
There is actually a sufficient and necessary condition for $M_n(mathbb{C})$:
Let $J$ be the Jordan canonical form of a complex matrix $A$, i.e.,
$$
A=PJP^{-1}=Pmathrm{diag}(J_1,cdots,J_s)P^{-1}
$$
where
$$
J_i=lambda_iI+N_i=left(begin{matrix}
lambda_i & & &\
1 & ddots & & \
& ddots & ddots & \
& & 1 & lambda_i
end{matrix}right).
$$
Then the matrices commutable with $A$ have the form of
$$
B=PB_1P^{-1}=P(B_{ij})P^{-1}
$$
where $B_1=(B_{ij})$ has the same blocking as $J$, and
$$
B_{ij}=begin{cases}
0 &mbox{if } lambda_inelambda_j\
mbox{a }unicode{x201C}mbox{lower-triangle-layered matrix''} & mbox{if } lambda_i=lambda_j
end{cases}
$$
The reference is my textbook of linear algebra, though not in English. Here is an example in the book:
The main part in the proof is to compare the corresponding entries in both sides of the equation $N_iB_{ij}=B_{ij}N_j$ when $lambda_i=lambda_j$.
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
$endgroup$
There is actually a sufficient and necessary condition for $M_n(mathbb{C})$:
Let $J$ be the Jordan canonical form of a complex matrix $A$, i.e.,
$$
A=PJP^{-1}=Pmathrm{diag}(J_1,cdots,J_s)P^{-1}
$$
where
$$
J_i=lambda_iI+N_i=left(begin{matrix}
lambda_i & & &\
1 & ddots & & \
& ddots & ddots & \
& & 1 & lambda_i
end{matrix}right).
$$
Then the matrices commutable with $A$ have the form of
$$
B=PB_1P^{-1}=P(B_{ij})P^{-1}
$$
where $B_1=(B_{ij})$ has the same blocking as $J$, and
$$
B_{ij}=begin{cases}
0 &mbox{if } lambda_inelambda_j\
mbox{a }unicode{x201C}mbox{lower-triangle-layered matrix''} & mbox{if } lambda_i=lambda_j
end{cases}
$$
The reference is my textbook of linear algebra, though not in English. Here is an example in the book:
The main part in the proof is to compare the corresponding entries in both sides of the equation $N_iB_{ij}=B_{ij}N_j$ when $lambda_i=lambda_j$.
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
edited Jan 4 at 11:56
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
answered Nov 15 '14 at 23:54
ziyuangziyuang
1,3151826
1,3151826
add a comment |
add a comment |
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5
$begingroup$
en.wikipedia.org/wiki/Commuting_matrices
$endgroup$
– B0rk4
Aug 29 '13 at 7:30
4
$begingroup$
Since everybody (except Hauke) is just listing their favorite sufficient conditions let me add mine: If there exists a polynomial $Pin R[X]$ ($R$ a commutative ring containing the entries of $A$ and $B$) such that $B=P(A)$, then we have $AB=BA$. Furthermore, in the case that $R$ is (contained in) an algebraically closed field and the eigenvalues of $A$ are distinct, then this sufficient criterion is also necessary. For more read the Wikiarticle linked to by Hauke.
$endgroup$
– Jyrki Lahtonen
Aug 29 '13 at 8:17