Is matrix multiplication transitive? [duplicate]

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This question already has an answer here:

• 
  If $A$ commutes with $B$ and $B$ commutes with $C$, then $A$ commutes with $C$
  
  1 answer
  
  

Let $A, B, C in M_n(mathbb R)$ be such that $A$ commutes with $B$, $B$ commutes with $C$ and $B$ is not a scalar matrix. Then $A$ commutes with $C$.

I think it is false, but how can I solve it within 3 minutes?

linear-algebra matrices

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edited Jul 22 at 20:44

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 15:38

Cloud JR

469414

marked as duplicate by Yanko, mechanodroid, Robert Soupe, Brevan Ellefsen, Parcly Taxel Jul 23 at 7:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Let $B$ be the identity matrix?
  – Lord Shark the Unknown
  Jul 22 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  B is not a scalar matrix ...btw identity is also a scalr matrix
  – Cloud JR
  Jul 22 at 15:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is a non-scalar matrix?
  – rwbogl
  Jul 22 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It means matrix B is not a scalar matrix
  – Cloud JR
  Jul 22 at 15:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By scalar matrix do you mean a matrix of the form $lambda I$ for some real $lambda$ and $I$ the identity matrix?
  – rwbogl
  Jul 22 at 15:51
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

2

This question already has an answer here:

• 
  If $A$ commutes with $B$ and $B$ commutes with $C$, then $A$ commutes with $C$
  
  1 answer
  
  

Let $A, B, C in M_n(mathbb R)$ be such that $A$ commutes with $B$, $B$ commutes with $C$ and $B$ is not a scalar matrix. Then $A$ commutes with $C$.

I think it is false, but how can I solve it within 3 minutes?

linear-algebra matrices

share|cite|improve this question

edited Jul 22 at 20:44

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 15:38

Cloud JR

469414

marked as duplicate by Yanko, mechanodroid, Robert Soupe, Brevan Ellefsen, Parcly Taxel Jul 23 at 7:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Let $B$ be the identity matrix?
  – Lord Shark the Unknown
  Jul 22 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  B is not a scalar matrix ...btw identity is also a scalr matrix
  – Cloud JR
  Jul 22 at 15:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is a non-scalar matrix?
  – rwbogl
  Jul 22 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It means matrix B is not a scalar matrix
  – Cloud JR
  Jul 22 at 15:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By scalar matrix do you mean a matrix of the form $lambda I$ for some real $lambda$ and $I$ the identity matrix?
  – rwbogl
  Jul 22 at 15:51
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

2

up vote
0
down vote

favorite

2

2

This question already has an answer here:

• 
  If $A$ commutes with $B$ and $B$ commutes with $C$, then $A$ commutes with $C$
  
  1 answer
  
  

Let $A, B, C in M_n(mathbb R)$ be such that $A$ commutes with $B$, $B$ commutes with $C$ and $B$ is not a scalar matrix. Then $A$ commutes with $C$.

I think it is false, but how can I solve it within 3 minutes?

linear-algebra matrices

share|cite|improve this question

edited Jul 22 at 20:44

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 15:38

Cloud JR

469414

This question already has an answer here:

• 
  If $A$ commutes with $B$ and $B$ commutes with $C$, then $A$ commutes with $C$
  
  1 answer
  
  

Let $A, B, C in M_n(mathbb R)$ be such that $A$ commutes with $B$, $B$ commutes with $C$ and $B$ is not a scalar matrix. Then $A$ commutes with $C$.

I think it is false, but how can I solve it within 3 minutes?

This question already has an answer here:

• 
  If $A$ commutes with $B$ and $B$ commutes with $C$, then $A$ commutes with $C$
  
  1 answer
  
  

linear-algebra matrices

share|cite|improve this question

edited Jul 22 at 20:44

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 15:38

Cloud JR

469414

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 20:44

Rodrigo de Azevedo

12.6k41751

edited Jul 22 at 20:44

Rodrigo de Azevedo

12.6k41751

edited Jul 22 at 20:44

Rodrigo de Azevedo

12.6k41751

12.6k41751

asked Jul 22 at 15:38

Cloud JR

469414

asked Jul 22 at 15:38

Cloud JR

469414

asked Jul 22 at 15:38

Cloud JR

469414

469414

marked as duplicate by Yanko, mechanodroid, Robert Soupe, Brevan Ellefsen, Parcly Taxel Jul 23 at 7:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Yanko, mechanodroid, Robert Soupe, Brevan Ellefsen, Parcly Taxel Jul 23 at 7:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Let $B$ be the identity matrix?
  – Lord Shark the Unknown
  Jul 22 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  B is not a scalar matrix ...btw identity is also a scalr matrix
  – Cloud JR
  Jul 22 at 15:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is a non-scalar matrix?
  – rwbogl
  Jul 22 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It means matrix B is not a scalar matrix
  – Cloud JR
  Jul 22 at 15:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By scalar matrix do you mean a matrix of the form $lambda I$ for some real $lambda$ and $I$ the identity matrix?
  – rwbogl
  Jul 22 at 15:51
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Let $B$ be the identity matrix?
  – Lord Shark the Unknown
  Jul 22 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  B is not a scalar matrix ...btw identity is also a scalr matrix
  – Cloud JR
  Jul 22 at 15:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is a non-scalar matrix?
  – rwbogl
  Jul 22 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It means matrix B is not a scalar matrix
  – Cloud JR
  Jul 22 at 15:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By scalar matrix do you mean a matrix of the form $lambda I$ for some real $lambda$ and $I$ the identity matrix?
  – rwbogl
  Jul 22 at 15:51
  

  
  
  
  

4

4

Let $B$ be the identity matrix?
– Lord Shark the Unknown
Jul 22 at 15:41

Let $B$ be the identity matrix?
– Lord Shark the Unknown
Jul 22 at 15:41

B is not a scalar matrix ...btw identity is also a scalr matrix
– Cloud JR
Jul 22 at 15:43

B is not a scalar matrix ...btw identity is also a scalr matrix
– Cloud JR
Jul 22 at 15:43

1

1

What is a non-scalar matrix?
– rwbogl
Jul 22 at 15:45

What is a non-scalar matrix?
– rwbogl
Jul 22 at 15:45

It means matrix B is not a scalar matrix
– Cloud JR
Jul 22 at 15:47

It means matrix B is not a scalar matrix
– Cloud JR
Jul 22 at 15:47

By scalar matrix do you mean a matrix of the form $lambda I$ for some real $lambda$ and $I$ the identity matrix?
– rwbogl
Jul 22 at 15:51

By scalar matrix do you mean a matrix of the form $lambda I$ for some real $lambda$ and $I$ the identity matrix?
– rwbogl
Jul 22 at 15:51

 | 
show 1 more comment

3 Answers
3

active

oldest

votes

up vote
5
down vote



accepted

In $mathbb R^5$, let $A$ be the matrix for which switches the first two standard basis vectors and preserves the others. That is, $Ae_1=e_2,Ae_2=e_1,$ and $Ae_i=e_i$ for $i=3,4,5$.

Let $C$ be the matrix which switches $e_2$ and $e_3$ (preserving the others), and let $B$ be the matrix which switches $e_4$ and $e_5$ (preserving the other). We have $A$ commutes with $B$, and $B$ with $C$, but not $A$ with $C$.

share|cite|improve this answer

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good insight,this is really helpful
  – Cloud JR
  Jul 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Though jose answer is good, i accept yours because i like the way you look at this question . Thanks
  – Cloud JR
  Jul 22 at 15:54
  

  
  
  
  

add a comment | 

up vote
5
down vote

Take any two non-commuting $2times2$ matrices$$beginpmatrixa&b\c&dendpmatrixtext and beginpmatrixa'&b'\c'&d'endpmatrix.$$Now, take$$A=beginpmatrixa&b&0\c&d&0\0&0&0endpmatrix, B=beginpmatrix0&0&0\0&0&0\0&0&1endpmatrixtext and C=beginpmatrixa'&b'&0\c'&d'&0\0&0&0endpmatrix.$$

share|cite|improve this answer

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate this
  – Cloud JR
  Jul 22 at 15:50
  

  
  
  
  

add a comment | 

up vote
2
down vote

Take $A,B,C$ block matrices of the form
$$ A = beginpmatrix A' & 0 \
0 & 0 endpmatrix, quad B = beginpmatrix 0 & 0 \
0 & B' endpmatrix, quad C = beginpmatrix C' & 0 \
0 & 0 endpmatrix, $$
where $A',C'$ are $k times k$, $B'$ is $(n-k)times (n-k)$, and $A'C'-C'A' neq 0$. Then $AB=BA=CA=AC=0$, but
$$AC-CA = beginpmatrix A'C'-C'A' & 0 \
0 & 0 endpmatrix neq 0. $$

share|cite|improve this answer

answered Jul 22 at 15:50

Chappers

55k74190

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

In $mathbb R^5$, let $A$ be the matrix for which switches the first two standard basis vectors and preserves the others. That is, $Ae_1=e_2,Ae_2=e_1,$ and $Ae_i=e_i$ for $i=3,4,5$.

Let $C$ be the matrix which switches $e_2$ and $e_3$ (preserving the others), and let $B$ be the matrix which switches $e_4$ and $e_5$ (preserving the other). We have $A$ commutes with $B$, and $B$ with $C$, but not $A$ with $C$.

share|cite|improve this answer

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good insight,this is really helpful
  – Cloud JR
  Jul 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Though jose answer is good, i accept yours because i like the way you look at this question . Thanks
  – Cloud JR
  Jul 22 at 15:54
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

In $mathbb R^5$, let $A$ be the matrix for which switches the first two standard basis vectors and preserves the others. That is, $Ae_1=e_2,Ae_2=e_1,$ and $Ae_i=e_i$ for $i=3,4,5$.

Let $C$ be the matrix which switches $e_2$ and $e_3$ (preserving the others), and let $B$ be the matrix which switches $e_4$ and $e_5$ (preserving the other). We have $A$ commutes with $B$, and $B$ with $C$, but not $A$ with $C$.

share|cite|improve this answer

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good insight,this is really helpful
  – Cloud JR
  Jul 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Though jose answer is good, i accept yours because i like the way you look at this question . Thanks
  – Cloud JR
  Jul 22 at 15:54
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

In $mathbb R^5$, let $A$ be the matrix for which switches the first two standard basis vectors and preserves the others. That is, $Ae_1=e_2,Ae_2=e_1,$ and $Ae_i=e_i$ for $i=3,4,5$.

Let $C$ be the matrix which switches $e_2$ and $e_3$ (preserving the others), and let $B$ be the matrix which switches $e_4$ and $e_5$ (preserving the other). We have $A$ commutes with $B$, and $B$ with $C$, but not $A$ with $C$.

share|cite|improve this answer

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

In $mathbb R^5$, let $A$ be the matrix for which switches the first two standard basis vectors and preserves the others. That is, $Ae_1=e_2,Ae_2=e_1,$ and $Ae_i=e_i$ for $i=3,4,5$.

Let $C$ be the matrix which switches $e_2$ and $e_3$ (preserving the others), and let $B$ be the matrix which switches $e_4$ and $e_5$ (preserving the other). We have $A$ commutes with $B$, and $B$ with $C$, but not $A$ with $C$.

share|cite|improve this answer

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

answered Jul 22 at 15:49

Mike Earnest

15.2k11644

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good insight,this is really helpful
  – Cloud JR
  Jul 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Though jose answer is good, i accept yours because i like the way you look at this question . Thanks
  – Cloud JR
  Jul 22 at 15:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good insight,this is really helpful
  – Cloud JR
  Jul 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Though jose answer is good, i accept yours because i like the way you look at this question . Thanks
  – Cloud JR
  Jul 22 at 15:54
  

  
  
  
  

Good insight,this is really helpful
– Cloud JR
Jul 22 at 15:53

Good insight,this is really helpful
– Cloud JR
Jul 22 at 15:53

Though jose answer is good, i accept yours because i like the way you look at this question . Thanks
– Cloud JR
Jul 22 at 15:54

Though jose answer is good, i accept yours because i like the way you look at this question . Thanks
– Cloud JR
Jul 22 at 15:54

add a comment | 

up vote
5
down vote

Take any two non-commuting $2times2$ matrices$$beginpmatrixa&b\c&dendpmatrixtext and beginpmatrixa'&b'\c'&d'endpmatrix.$$Now, take$$A=beginpmatrixa&b&0\c&d&0\0&0&0endpmatrix, B=beginpmatrix0&0&0\0&0&0\0&0&1endpmatrixtext and C=beginpmatrixa'&b'&0\c'&d'&0\0&0&0endpmatrix.$$

share|cite|improve this answer

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate this
  – Cloud JR
  Jul 22 at 15:50
  

  
  
  
  

add a comment | 

up vote
5
down vote

Take any two non-commuting $2times2$ matrices$$beginpmatrixa&b\c&dendpmatrixtext and beginpmatrixa'&b'\c'&d'endpmatrix.$$Now, take$$A=beginpmatrixa&b&0\c&d&0\0&0&0endpmatrix, B=beginpmatrix0&0&0\0&0&0\0&0&1endpmatrixtext and C=beginpmatrixa'&b'&0\c'&d'&0\0&0&0endpmatrix.$$

share|cite|improve this answer

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate this
  – Cloud JR
  Jul 22 at 15:50
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

Take any two non-commuting $2times2$ matrices$$beginpmatrixa&b\c&dendpmatrixtext and beginpmatrixa'&b'\c'&d'endpmatrix.$$Now, take$$A=beginpmatrixa&b&0\c&d&0\0&0&0endpmatrix, B=beginpmatrix0&0&0\0&0&0\0&0&1endpmatrixtext and C=beginpmatrixa'&b'&0\c'&d'&0\0&0&0endpmatrix.$$

share|cite|improve this answer

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

Take any two non-commuting $2times2$ matrices$$beginpmatrixa&b\c&dendpmatrixtext and beginpmatrixa'&b'\c'&d'endpmatrix.$$Now, take$$A=beginpmatrixa&b&0\c&d&0\0&0&0endpmatrix, B=beginpmatrix0&0&0\0&0&0\0&0&1endpmatrixtext and C=beginpmatrixa'&b'&0\c'&d'&0\0&0&0endpmatrix.$$

share|cite|improve this answer

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

answered Jul 22 at 15:47

José Carlos Santos

113k1698176

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate this
  – Cloud JR
  Jul 22 at 15:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate this
  – Cloud JR
  Jul 22 at 15:50
  

  
  
  
  

I appreciate this
– Cloud JR
Jul 22 at 15:50

I appreciate this
– Cloud JR
Jul 22 at 15:50

add a comment | 

up vote
2
down vote

Take $A,B,C$ block matrices of the form
$$ A = beginpmatrix A' & 0 \
0 & 0 endpmatrix, quad B = beginpmatrix 0 & 0 \
0 & B' endpmatrix, quad C = beginpmatrix C' & 0 \
0 & 0 endpmatrix, $$
where $A',C'$ are $k times k$, $B'$ is $(n-k)times (n-k)$, and $A'C'-C'A' neq 0$. Then $AB=BA=CA=AC=0$, but
$$AC-CA = beginpmatrix A'C'-C'A' & 0 \
0 & 0 endpmatrix neq 0. $$

share|cite|improve this answer

answered Jul 22 at 15:50

Chappers

55k74190

add a comment | 

up vote
2
down vote

Take $A,B,C$ block matrices of the form
$$ A = beginpmatrix A' & 0 \
0 & 0 endpmatrix, quad B = beginpmatrix 0 & 0 \
0 & B' endpmatrix, quad C = beginpmatrix C' & 0 \
0 & 0 endpmatrix, $$
where $A',C'$ are $k times k$, $B'$ is $(n-k)times (n-k)$, and $A'C'-C'A' neq 0$. Then $AB=BA=CA=AC=0$, but
$$AC-CA = beginpmatrix A'C'-C'A' & 0 \
0 & 0 endpmatrix neq 0. $$

share|cite|improve this answer

answered Jul 22 at 15:50

Chappers

55k74190

add a comment | 

up vote
2
down vote

up vote
2
down vote

Take $A,B,C$ block matrices of the form
$$ A = beginpmatrix A' & 0 \
0 & 0 endpmatrix, quad B = beginpmatrix 0 & 0 \
0 & B' endpmatrix, quad C = beginpmatrix C' & 0 \
0 & 0 endpmatrix, $$
where $A',C'$ are $k times k$, $B'$ is $(n-k)times (n-k)$, and $A'C'-C'A' neq 0$. Then $AB=BA=CA=AC=0$, but
$$AC-CA = beginpmatrix A'C'-C'A' & 0 \
0 & 0 endpmatrix neq 0. $$

share|cite|improve this answer

answered Jul 22 at 15:50

Chappers

55k74190

Take $A,B,C$ block matrices of the form
$$ A = beginpmatrix A' & 0 \
0 & 0 endpmatrix, quad B = beginpmatrix 0 & 0 \
0 & B' endpmatrix, quad C = beginpmatrix C' & 0 \
0 & 0 endpmatrix, $$
where $A',C'$ are $k times k$, $B'$ is $(n-k)times (n-k)$, and $A'C'-C'A' neq 0$. Then $AB=BA=CA=AC=0$, but
$$AC-CA = beginpmatrix A'C'-C'A' & 0 \
0 & 0 endpmatrix neq 0. $$

share|cite|improve this answer

answered Jul 22 at 15:50

Chappers

55k74190

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 15:50

Chappers

55k74190

answered Jul 22 at 15:50

Chappers

55k74190

answered Jul 22 at 15:50

Chappers

55k74190

55k74190

add a comment | 

add a comment |