If $A$ and $B$ are two matrices, and $AB=I$ (of suitable order) is $BA=I$ always true? [duplicate]

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  If $AB = I$ then $BA = I$
  
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It is said that a matrix $A$ is said to be invertible if there exists a matrix $B$ such that $AB=I=BA$, but if we only know $AB=I$, can we be sure that $BA=I$?

matrices

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edited Jul 17 at 7:07

TheSimpliFire

9,71061951

asked Jul 17 at 7:00

Shashwat

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marked as duplicate by José Carlos Santos, Alex Francisco, Mostafa Ayaz, Arthur, Dietrich Burde Jul 17 at 8:24

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.

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  If $A$ is square, then yes, always (see linked duplicate). If $A$ is not square, then no, never (look at the maximum ranks).
  – Arthur
  Jul 17 at 7:35
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

This question already has an answer here:

• 
  If $AB = I$ then $BA = I$
  
  25 answers
  
  

It is said that a matrix $A$ is said to be invertible if there exists a matrix $B$ such that $AB=I=BA$, but if we only know $AB=I$, can we be sure that $BA=I$?

matrices

share|cite|improve this question

edited Jul 17 at 7:07

TheSimpliFire

9,71061951

asked Jul 17 at 7:00

Shashwat

93

marked as duplicate by José Carlos Santos, Alex Francisco, Mostafa Ayaz, Arthur, Dietrich Burde Jul 17 at 8:24

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is square, then yes, always (see linked duplicate). If $A$ is not square, then no, never (look at the maximum ranks).
  – Arthur
  Jul 17 at 7:35
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

This question already has an answer here:

• 
  If $AB = I$ then $BA = I$
  
  25 answers
  
  

It is said that a matrix $A$ is said to be invertible if there exists a matrix $B$ such that $AB=I=BA$, but if we only know $AB=I$, can we be sure that $BA=I$?

matrices

share|cite|improve this question

edited Jul 17 at 7:07

TheSimpliFire

9,71061951

asked Jul 17 at 7:00

Shashwat

93

This question already has an answer here:

• 
  If $AB = I$ then $BA = I$
  
  25 answers
  
  

It is said that a matrix $A$ is said to be invertible if there exists a matrix $B$ such that $AB=I=BA$, but if we only know $AB=I$, can we be sure that $BA=I$?

This question already has an answer here:

• 
  If $AB = I$ then $BA = I$
  
  25 answers
  
  

matrices

share|cite|improve this question

edited Jul 17 at 7:07

TheSimpliFire

9,71061951

asked Jul 17 at 7:00

Shashwat

93

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 7:07

TheSimpliFire

9,71061951

edited Jul 17 at 7:07

TheSimpliFire

9,71061951

edited Jul 17 at 7:07

TheSimpliFire

9,71061951

9,71061951

asked Jul 17 at 7:00

Shashwat

93

asked Jul 17 at 7:00

Shashwat

93

asked Jul 17 at 7:00

Shashwat

93

93

marked as duplicate by José Carlos Santos, Alex Francisco, Mostafa Ayaz, Arthur, Dietrich Burde Jul 17 at 8:24

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 José Carlos Santos, Alex Francisco, Mostafa Ayaz, Arthur, Dietrich Burde Jul 17 at 8:24

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is square, then yes, always (see linked duplicate). If $A$ is not square, then no, never (look at the maximum ranks).
  – Arthur
  Jul 17 at 7:35
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is square, then yes, always (see linked duplicate). If $A$ is not square, then no, never (look at the maximum ranks).
  – Arthur
  Jul 17 at 7:35
  
  

  
  
  
  

If $A$ is square, then yes, always (see linked duplicate). If $A$ is not square, then no, never (look at the maximum ranks).
– Arthur
Jul 17 at 7:35

If $A$ is square, then yes, always (see linked duplicate). If $A$ is not square, then no, never (look at the maximum ranks).
– Arthur
Jul 17 at 7:35

add a comment | 

2 Answers
2

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up vote
2
down vote

If $A$ is invertible then $$AB=Iimplies ABA=IA=Aimplies A^-1ABA=A^-1Aimplies BA=I.$$

share|cite|improve this answer

edited Jul 17 at 7:44

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  True if $A^-1$ exists. But what about the case $A=pmatrix1&0&0cr0&1&0cr$, $B=pmatrix1&0cr0&1cr0&0cr$?
  – David
  Jul 17 at 7:33
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

If your matrix are square, then is true. A proof is here. The proof doesn't suppose $A^-1$ exists. If not, for example
$$
A = beginpmatrix
2 & 0 & 1\
0 & 0 & 2
endpmatrixqquad
B = frac14beginpmatrix
2 & -1\
0 & 0 \
0 & 2
endpmatrix
$$
Then
$$
AB= I_2 qquad BA = beginpmatrix1& 0 &0\ 0 & 0 &0 \ 0& 0 & 1endpmatrix
$$

share|cite|improve this answer

edited Jul 17 at 10:06

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

If $A$ is invertible then $$AB=Iimplies ABA=IA=Aimplies A^-1ABA=A^-1Aimplies BA=I.$$

share|cite|improve this answer

edited Jul 17 at 7:44

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  True if $A^-1$ exists. But what about the case $A=pmatrix1&0&0cr0&1&0cr$, $B=pmatrix1&0cr0&1cr0&0cr$?
  – David
  Jul 17 at 7:33
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

If $A$ is invertible then $$AB=Iimplies ABA=IA=Aimplies A^-1ABA=A^-1Aimplies BA=I.$$

share|cite|improve this answer

edited Jul 17 at 7:44

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  True if $A^-1$ exists. But what about the case $A=pmatrix1&0&0cr0&1&0cr$, $B=pmatrix1&0cr0&1cr0&0cr$?
  – David
  Jul 17 at 7:33
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

If $A$ is invertible then $$AB=Iimplies ABA=IA=Aimplies A^-1ABA=A^-1Aimplies BA=I.$$

share|cite|improve this answer

edited Jul 17 at 7:44

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

If $A$ is invertible then $$AB=Iimplies ABA=IA=Aimplies A^-1ABA=A^-1Aimplies BA=I.$$

share|cite|improve this answer

edited Jul 17 at 7:44

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 7:44

edited Jul 17 at 7:44

edited Jul 17 at 7:44

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

answered Jul 17 at 7:06

TheSimpliFire

9,71061951

9,71061951

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  True if $A^-1$ exists. But what about the case $A=pmatrix1&0&0cr0&1&0cr$, $B=pmatrix1&0cr0&1cr0&0cr$?
  – David
  Jul 17 at 7:33
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  True if $A^-1$ exists. But what about the case $A=pmatrix1&0&0cr0&1&0cr$, $B=pmatrix1&0cr0&1cr0&0cr$?
  – David
  Jul 17 at 7:33
  
  

  
  
  
  

5

5

True if $A^-1$ exists. But what about the case $A=pmatrix1&0&0cr0&1&0cr$, $B=pmatrix1&0cr0&1cr0&0cr$?
– David
Jul 17 at 7:33

True if $A^-1$ exists. But what about the case $A=pmatrix1&0&0cr0&1&0cr$, $B=pmatrix1&0cr0&1cr0&0cr$?
– David
Jul 17 at 7:33

add a comment | 

up vote
2
down vote

If your matrix are square, then is true. A proof is here. The proof doesn't suppose $A^-1$ exists. If not, for example
$$
A = beginpmatrix
2 & 0 & 1\
0 & 0 & 2
endpmatrixqquad
B = frac14beginpmatrix
2 & -1\
0 & 0 \
0 & 2
endpmatrix
$$
Then
$$
AB= I_2 qquad BA = beginpmatrix1& 0 &0\ 0 & 0 &0 \ 0& 0 & 1endpmatrix
$$

share|cite|improve this answer

edited Jul 17 at 10:06

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

add a comment | 

up vote
2
down vote

If your matrix are square, then is true. A proof is here. The proof doesn't suppose $A^-1$ exists. If not, for example
$$
A = beginpmatrix
2 & 0 & 1\
0 & 0 & 2
endpmatrixqquad
B = frac14beginpmatrix
2 & -1\
0 & 0 \
0 & 2
endpmatrix
$$
Then
$$
AB= I_2 qquad BA = beginpmatrix1& 0 &0\ 0 & 0 &0 \ 0& 0 & 1endpmatrix
$$

share|cite|improve this answer

edited Jul 17 at 10:06

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

add a comment | 

up vote
2
down vote

up vote
2
down vote

If your matrix are square, then is true. A proof is here. The proof doesn't suppose $A^-1$ exists. If not, for example
$$
A = beginpmatrix
2 & 0 & 1\
0 & 0 & 2
endpmatrixqquad
B = frac14beginpmatrix
2 & -1\
0 & 0 \
0 & 2
endpmatrix
$$
Then
$$
AB= I_2 qquad BA = beginpmatrix1& 0 &0\ 0 & 0 &0 \ 0& 0 & 1endpmatrix
$$

share|cite|improve this answer

edited Jul 17 at 10:06

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

If your matrix are square, then is true. A proof is here. The proof doesn't suppose $A^-1$ exists. If not, for example
$$
A = beginpmatrix
2 & 0 & 1\
0 & 0 & 2
endpmatrixqquad
B = frac14beginpmatrix
2 & -1\
0 & 0 \
0 & 2
endpmatrix
$$
Then
$$
AB= I_2 qquad BA = beginpmatrix1& 0 &0\ 0 & 0 &0 \ 0& 0 & 1endpmatrix
$$

share|cite|improve this answer

edited Jul 17 at 10:06

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 10:06

edited Jul 17 at 10:06

edited Jul 17 at 10:06

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

answered Jul 17 at 8:13

Rafael Gonzalez Lopez

652112

652112

add a comment | 

add a comment |