If $A, B$ and $C$ are all non-zero matrices, and $AB = BC$, does it imply $A = C$?

0

$begingroup$

I don't see how $A = C$ but I can't find an example that proves otherwise.

matrices

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edited Jan 20 at 22:48

idriskameni

753321

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint : do you have heard about matrix reduction ? ! ...
  $endgroup$
  – DLeMeur
  Jan 20 at 22:08
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I don't see how $A = C$ but I can't find an example that proves otherwise.

matrices

share|cite|improve this question

edited Jan 20 at 22:48

idriskameni

753321

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint : do you have heard about matrix reduction ? ! ...
  $endgroup$
  – DLeMeur
  Jan 20 at 22:08
  
  
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

I don't see how $A = C$ but I can't find an example that proves otherwise.

matrices

share|cite|improve this question

edited Jan 20 at 22:48

idriskameni

753321

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

$endgroup$

I don't see how $A = C$ but I can't find an example that proves otherwise.

matrices

matrices

share|cite|improve this question

edited Jan 20 at 22:48

idriskameni

753321

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

share|cite|improve this question

edited Jan 20 at 22:48

idriskameni

753321

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

share|cite|improve this question

share|cite|improve this question

edited Jan 20 at 22:48

idriskameni

753321

edited Jan 20 at 22:48

idriskameni

753321

edited Jan 20 at 22:48

idriskameni

753321

753321

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

asked Jan 20 at 22:03

Lim Chee HeanLim Chee Hean

6

6

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint : do you have heard about matrix reduction ? ! ...
  $endgroup$
  – DLeMeur
  Jan 20 at 22:08
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint : do you have heard about matrix reduction ? ! ...
  $endgroup$
  – DLeMeur
  Jan 20 at 22:08
  
  
  

  
  
  
  

$begingroup$
Hint : do you have heard about matrix reduction ? ! ...
$endgroup$
– DLeMeur
Jan 20 at 22:08

$begingroup$
Hint : do you have heard about matrix reduction ? ! ...
$endgroup$
– DLeMeur
Jan 20 at 22:08

add a comment | 

3 Answers
3

active

oldest

votes

4

$begingroup$

$$A=begin {pmatrix} 1&0\0&1 end {pmatrix},B=begin {pmatrix} 1&0\0&0 end {pmatrix},C=begin {pmatrix} 1&0\0&0 end {pmatrix}$$

In fact, you should be able to use generic matrices for this. If $B$ is invertible and does not commute with $A$ you can just set $C=B^{-1}AB$ and have an example. We can let
$$B=begin {pmatrix} 1&2\1&3 end {pmatrix},B^{-1}=begin {pmatrix} 3&-2\-1&1 end {pmatrix},A=begin {pmatrix} 1&2\3&4 end {pmatrix},C=begin {pmatrix} -5&-12\4&10 end {pmatrix}$$

share|cite|improve this answer

edited Jan 20 at 23:20

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

$endgroup$

add a comment | 

1

$begingroup$

$$A=begin {pmatrix} 1&1\1&2 end {pmatrix},B=begin {pmatrix} 1&0\1&1 end {pmatrix},C=begin {pmatrix} 2&1\1&1 end {pmatrix}$$

share|cite|improve this answer

answered Jan 20 at 22:13

idriskameniidriskameni

753321

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate that you've chosen an example where the matrices are invertible
  $endgroup$
  – Omnomnomnom
  Jan 20 at 22:24
  

  
  
  
  

add a comment | 

1

$begingroup$

Matrix multiplication is generally not commutative, your proposition is false. When cooking up counter examples, you should look to use matrices with very basic properties. @ross-millikan has provided a good example of this.

It is very useful to consider the Identity matrix, Zero matrix (for other examples) and then matrices consisting of a reordering of the columns of the identity matrix.

Ross' examples can be written as

$$A=I_{2text{x}2}, quad B = C = e_1 e_1^T$$

where $e_i$ represents the $i$th column of the Identity matrix.

share|cite|improve this answer

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

4

$begingroup$

$$A=begin {pmatrix} 1&0\0&1 end {pmatrix},B=begin {pmatrix} 1&0\0&0 end {pmatrix},C=begin {pmatrix} 1&0\0&0 end {pmatrix}$$

In fact, you should be able to use generic matrices for this. If $B$ is invertible and does not commute with $A$ you can just set $C=B^{-1}AB$ and have an example. We can let
$$B=begin {pmatrix} 1&2\1&3 end {pmatrix},B^{-1}=begin {pmatrix} 3&-2\-1&1 end {pmatrix},A=begin {pmatrix} 1&2\3&4 end {pmatrix},C=begin {pmatrix} -5&-12\4&10 end {pmatrix}$$

share|cite|improve this answer

edited Jan 20 at 23:20

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

$endgroup$

add a comment | 

4

$begingroup$

$$A=begin {pmatrix} 1&0\0&1 end {pmatrix},B=begin {pmatrix} 1&0\0&0 end {pmatrix},C=begin {pmatrix} 1&0\0&0 end {pmatrix}$$

In fact, you should be able to use generic matrices for this. If $B$ is invertible and does not commute with $A$ you can just set $C=B^{-1}AB$ and have an example. We can let
$$B=begin {pmatrix} 1&2\1&3 end {pmatrix},B^{-1}=begin {pmatrix} 3&-2\-1&1 end {pmatrix},A=begin {pmatrix} 1&2\3&4 end {pmatrix},C=begin {pmatrix} -5&-12\4&10 end {pmatrix}$$

share|cite|improve this answer

edited Jan 20 at 23:20

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

$endgroup$

add a comment | 

4

4

4

$begingroup$

$$A=begin {pmatrix} 1&0\0&1 end {pmatrix},B=begin {pmatrix} 1&0\0&0 end {pmatrix},C=begin {pmatrix} 1&0\0&0 end {pmatrix}$$

In fact, you should be able to use generic matrices for this. If $B$ is invertible and does not commute with $A$ you can just set $C=B^{-1}AB$ and have an example. We can let
$$B=begin {pmatrix} 1&2\1&3 end {pmatrix},B^{-1}=begin {pmatrix} 3&-2\-1&1 end {pmatrix},A=begin {pmatrix} 1&2\3&4 end {pmatrix},C=begin {pmatrix} -5&-12\4&10 end {pmatrix}$$

share|cite|improve this answer

edited Jan 20 at 23:20

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

$endgroup$

$$A=begin {pmatrix} 1&0\0&1 end {pmatrix},B=begin {pmatrix} 1&0\0&0 end {pmatrix},C=begin {pmatrix} 1&0\0&0 end {pmatrix}$$

In fact, you should be able to use generic matrices for this. If $B$ is invertible and does not commute with $A$ you can just set $C=B^{-1}AB$ and have an example. We can let
$$B=begin {pmatrix} 1&2\1&3 end {pmatrix},B^{-1}=begin {pmatrix} 3&-2\-1&1 end {pmatrix},A=begin {pmatrix} 1&2\3&4 end {pmatrix},C=begin {pmatrix} -5&-12\4&10 end {pmatrix}$$

share|cite|improve this answer

edited Jan 20 at 23:20

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

share|cite|improve this answer

share|cite|improve this answer

edited Jan 20 at 23:20

edited Jan 20 at 23:20

edited Jan 20 at 23:20

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

answered Jan 20 at 22:10

Ross MillikanRoss Millikan

298k24200373

298k24200373

add a comment | 

add a comment | 

1

$begingroup$

$$A=begin {pmatrix} 1&1\1&2 end {pmatrix},B=begin {pmatrix} 1&0\1&1 end {pmatrix},C=begin {pmatrix} 2&1\1&1 end {pmatrix}$$

share|cite|improve this answer

answered Jan 20 at 22:13

idriskameniidriskameni

753321

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate that you've chosen an example where the matrices are invertible
  $endgroup$
  – Omnomnomnom
  Jan 20 at 22:24
  

  
  
  
  

add a comment | 

1

$begingroup$

$$A=begin {pmatrix} 1&1\1&2 end {pmatrix},B=begin {pmatrix} 1&0\1&1 end {pmatrix},C=begin {pmatrix} 2&1\1&1 end {pmatrix}$$

share|cite|improve this answer

answered Jan 20 at 22:13

idriskameniidriskameni

753321

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate that you've chosen an example where the matrices are invertible
  $endgroup$
  – Omnomnomnom
  Jan 20 at 22:24
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

$$A=begin {pmatrix} 1&1\1&2 end {pmatrix},B=begin {pmatrix} 1&0\1&1 end {pmatrix},C=begin {pmatrix} 2&1\1&1 end {pmatrix}$$

share|cite|improve this answer

answered Jan 20 at 22:13

idriskameniidriskameni

753321

$endgroup$

$$A=begin {pmatrix} 1&1\1&2 end {pmatrix},B=begin {pmatrix} 1&0\1&1 end {pmatrix},C=begin {pmatrix} 2&1\1&1 end {pmatrix}$$

share|cite|improve this answer

answered Jan 20 at 22:13

idriskameniidriskameni

753321

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 22:13

idriskameniidriskameni

753321

answered Jan 20 at 22:13

idriskameniidriskameni

753321

answered Jan 20 at 22:13

idriskameniidriskameni

753321

753321

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate that you've chosen an example where the matrices are invertible
  $endgroup$
  – Omnomnomnom
  Jan 20 at 22:24
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate that you've chosen an example where the matrices are invertible
  $endgroup$
  – Omnomnomnom
  Jan 20 at 22:24
  

  
  
  
  

1

1

$begingroup$
I appreciate that you've chosen an example where the matrices are invertible
$endgroup$
– Omnomnomnom
Jan 20 at 22:24

$begingroup$
I appreciate that you've chosen an example where the matrices are invertible
$endgroup$
– Omnomnomnom
Jan 20 at 22:24

add a comment | 

1

$begingroup$

Matrix multiplication is generally not commutative, your proposition is false. When cooking up counter examples, you should look to use matrices with very basic properties. @ross-millikan has provided a good example of this.

It is very useful to consider the Identity matrix, Zero matrix (for other examples) and then matrices consisting of a reordering of the columns of the identity matrix.

Ross' examples can be written as

$$A=I_{2text{x}2}, quad B = C = e_1 e_1^T$$

where $e_i$ represents the $i$th column of the Identity matrix.

share|cite|improve this answer

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

$endgroup$

add a comment | 

1

$begingroup$

Matrix multiplication is generally not commutative, your proposition is false. When cooking up counter examples, you should look to use matrices with very basic properties. @ross-millikan has provided a good example of this.

It is very useful to consider the Identity matrix, Zero matrix (for other examples) and then matrices consisting of a reordering of the columns of the identity matrix.

Ross' examples can be written as

$$A=I_{2text{x}2}, quad B = C = e_1 e_1^T$$

where $e_i$ represents the $i$th column of the Identity matrix.

share|cite|improve this answer

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

$endgroup$

add a comment | 

1

1

1

$begingroup$

Matrix multiplication is generally not commutative, your proposition is false. When cooking up counter examples, you should look to use matrices with very basic properties. @ross-millikan has provided a good example of this.

It is very useful to consider the Identity matrix, Zero matrix (for other examples) and then matrices consisting of a reordering of the columns of the identity matrix.

Ross' examples can be written as

$$A=I_{2text{x}2}, quad B = C = e_1 e_1^T$$

where $e_i$ represents the $i$th column of the Identity matrix.

share|cite|improve this answer

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

$endgroup$

Matrix multiplication is generally not commutative, your proposition is false. When cooking up counter examples, you should look to use matrices with very basic properties. @ross-millikan has provided a good example of this.

It is very useful to consider the Identity matrix, Zero matrix (for other examples) and then matrices consisting of a reordering of the columns of the identity matrix.

Ross' examples can be written as

$$A=I_{2text{x}2}, quad B = C = e_1 e_1^T$$

where $e_i$ represents the $i$th column of the Identity matrix.

share|cite|improve this answer

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

answered Jan 20 at 22:16

Joel BiffinJoel Biffin

1017

1017

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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