If $A+B=AB$ then $AB=BA$












9












$begingroup$


I was doing the problem $$ A+B=ABimplies AB=BA. $$



$AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:23






  • 5




    $begingroup$
    The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^{-1}$ exists).
    $endgroup$
    – user228113
    Apr 19 '17 at 23:26












  • $begingroup$
    @Coffeemath Why not same
    $endgroup$
    – Sachchidanand Prasad
    Apr 19 '17 at 23:29










  • $begingroup$
    If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:37






  • 3




    $begingroup$
    Just to say, suppose $A$ and $B$ are both the zero matrix. Then of course $A+B=AB=BA$ but neither $A$ nor $B$ is invertible.
    $endgroup$
    – lulu
    Apr 19 '17 at 23:38
















9












$begingroup$


I was doing the problem $$ A+B=ABimplies AB=BA. $$



$AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:23






  • 5




    $begingroup$
    The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^{-1}$ exists).
    $endgroup$
    – user228113
    Apr 19 '17 at 23:26












  • $begingroup$
    @Coffeemath Why not same
    $endgroup$
    – Sachchidanand Prasad
    Apr 19 '17 at 23:29










  • $begingroup$
    If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:37






  • 3




    $begingroup$
    Just to say, suppose $A$ and $B$ are both the zero matrix. Then of course $A+B=AB=BA$ but neither $A$ nor $B$ is invertible.
    $endgroup$
    – lulu
    Apr 19 '17 at 23:38














9












9








9


4



$begingroup$


I was doing the problem $$ A+B=ABimplies AB=BA. $$



$AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.






linear-algebra matrices







share|cite|improve this question











$endgroup$




I was doing the problem $$ A+B=ABimplies AB=BA. $$



$AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 11:24









greedoid

44k1155109




44k1155109










asked Apr 19 '17 at 23:20









Sachchidanand PrasadSachchidanand Prasad

1,674721




1,674721








  • 1




    $begingroup$
    Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:23






  • 5




    $begingroup$
    The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^{-1}$ exists).
    $endgroup$
    – user228113
    Apr 19 '17 at 23:26












  • $begingroup$
    @Coffeemath Why not same
    $endgroup$
    – Sachchidanand Prasad
    Apr 19 '17 at 23:29










  • $begingroup$
    If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:37






  • 3




    $begingroup$
    Just to say, suppose $A$ and $B$ are both the zero matrix. Then of course $A+B=AB=BA$ but neither $A$ nor $B$ is invertible.
    $endgroup$
    – lulu
    Apr 19 '17 at 23:38














  • 1




    $begingroup$
    Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:23






  • 5




    $begingroup$
    The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^{-1}$ exists).
    $endgroup$
    – user228113
    Apr 19 '17 at 23:26












  • $begingroup$
    @Coffeemath Why not same
    $endgroup$
    – Sachchidanand Prasad
    Apr 19 '17 at 23:29










  • $begingroup$
    If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA.
    $endgroup$
    – coffeemath
    Apr 19 '17 at 23:37






  • 3




    $begingroup$
    Just to say, suppose $A$ and $B$ are both the zero matrix. Then of course $A+B=AB=BA$ but neither $A$ nor $B$ is invertible.
    $endgroup$
    – lulu
    Apr 19 '17 at 23:38








1




1




$begingroup$
Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible.
$endgroup$
– coffeemath
Apr 19 '17 at 23:23




$begingroup$
Note that $AB=BA$ is not the same, nor does it imply, that $A,B$ are invertible.
$endgroup$
– coffeemath
Apr 19 '17 at 23:23




5




5




$begingroup$
The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^{-1}$ exists).
$endgroup$
– user228113
Apr 19 '17 at 23:26






$begingroup$
The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^{-1}$ exists).
$endgroup$
– user228113
Apr 19 '17 at 23:26














$begingroup$
@Coffeemath Why not same
$endgroup$
– Sachchidanand Prasad
Apr 19 '17 at 23:29




$begingroup$
@Coffeemath Why not same
$endgroup$
– Sachchidanand Prasad
Apr 19 '17 at 23:29












$begingroup$
If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA.
$endgroup$
– coffeemath
Apr 19 '17 at 23:37




$begingroup$
If A and B are equal, where each has rows [0,1],[0,0] then these aren't invertible, even though AB=BA.
$endgroup$
– coffeemath
Apr 19 '17 at 23:37




3




3




$begingroup$
Just to say, suppose $A$ and $B$ are both the zero matrix. Then of course $A+B=AB=BA$ but neither $A$ nor $B$ is invertible.
$endgroup$
– lulu
Apr 19 '17 at 23:38




$begingroup$
Just to say, suppose $A$ and $B$ are both the zero matrix. Then of course $A+B=AB=BA$ but neither $A$ nor $B$ is invertible.
$endgroup$
– lulu
Apr 19 '17 at 23:38










2 Answers
2






active

oldest

votes


















34












$begingroup$

Consider the expression $$(A-mathbb 1)(B-mathbb 1)=AB-A-B+mathbb 1=mathbb 1$$



Thus $(A-mathbb 1)$ and $(B-mathbb 1)$ are inverse to each other, whence $$mathbb 1= (B-mathbb 1)(A-mathbb 1)=BA - A - B + mathbb 1$$



It follows that $$BA=A+B=AB$$ and we are done.



Note: here $mathbb 1$ denotes the appropriate identity matrix.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if $ a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ?
    $endgroup$
    – освящение
    Jan 19 '18 at 4:37












  • $begingroup$
    If yes, then please explain!
    $endgroup$
    – освящение
    Jan 19 '18 at 4:39






  • 1




    $begingroup$
    Yes, $X,Y$ are invertible matrices and these form a group.
    $endgroup$
    – lulu
    Jan 19 '18 at 10:47










  • $begingroup$
    but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$
    $endgroup$
    – освящение
    Jan 19 '18 at 15:10






  • 1




    $begingroup$
    @Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=mathbb 1$. If nothing else, that equation shows that $det X times det Y =1neq 0$ so both $X,Y$ have to be invertible,
    $endgroup$
    – lulu
    Jan 19 '18 at 15:14



















1












$begingroup$

Note: $I$ is Identity Matrix



Start with the equation given by the problem:
$$A+B=AB$$
left multiply both sides of this equation by B that produces:
$$B(A+B)=BAB$$
$$BA+BB=BAB$$
$$BA+BB=(BA)B$$
$$BB=(BA)B - BA$$
$$BB=(BA)(B - I);;(1)$$



then right multiply both sides of this equation by B that produces:



$$(A+B)B=ABB$$
$$AB+BB=ABB$$
$$AB+BB=(AB)B$$
$$BB=(AB)B - AB$$
$$BB=(AB)(B - I);;(2)$$



and we subtract equation(1) from equation(2) that gives:
$$0 = (AB - BA)(B - I)$$
if $B = I$,the equation $A+B=AB$ is broken. So, we can know $(AB - BA) = 0$ that means $$AB = BA$$



The purpose of this problem is to find the BA combination, we can get it by left multiplying.



Sorry,
if $B - I$ is a singular matrix, then $AB - BA$ may not be $0$.
So, I will try to find $M(B - I) = I$,
as above, start with equation:
$$A+B=AB$$
replace $B$ with $(B - I) + I$ in left side
$$A+(B - I) + I =AB$$
$$I =AB - A - (B - I)$$
$$I =A(B - I) - (B - I)$$
$$I =(A - I)(B - I)$$
the inverse matrix of $(B - I)$ is $(A - I)$, so $(B - I)$ is invertible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=begin{bmatrix}1&0\0&0end{bmatrix}$ and $Y=begin{bmatrix}0&0\0&1end{bmatrix}$
    $endgroup$
    – obareey
    Dec 26 '18 at 12:53












  • $begingroup$
    @obareey thanks, I have re-edited
    $endgroup$
    – Revc_Ra
    Dec 26 '18 at 15:59











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









34












$begingroup$

Consider the expression $$(A-mathbb 1)(B-mathbb 1)=AB-A-B+mathbb 1=mathbb 1$$



Thus $(A-mathbb 1)$ and $(B-mathbb 1)$ are inverse to each other, whence $$mathbb 1= (B-mathbb 1)(A-mathbb 1)=BA - A - B + mathbb 1$$



It follows that $$BA=A+B=AB$$ and we are done.



Note: here $mathbb 1$ denotes the appropriate identity matrix.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if $ a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ?
    $endgroup$
    – освящение
    Jan 19 '18 at 4:37












  • $begingroup$
    If yes, then please explain!
    $endgroup$
    – освящение
    Jan 19 '18 at 4:39






  • 1




    $begingroup$
    Yes, $X,Y$ are invertible matrices and these form a group.
    $endgroup$
    – lulu
    Jan 19 '18 at 10:47










  • $begingroup$
    but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$
    $endgroup$
    – освящение
    Jan 19 '18 at 15:10






  • 1




    $begingroup$
    @Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=mathbb 1$. If nothing else, that equation shows that $det X times det Y =1neq 0$ so both $X,Y$ have to be invertible,
    $endgroup$
    – lulu
    Jan 19 '18 at 15:14
















34












$begingroup$

Consider the expression $$(A-mathbb 1)(B-mathbb 1)=AB-A-B+mathbb 1=mathbb 1$$



Thus $(A-mathbb 1)$ and $(B-mathbb 1)$ are inverse to each other, whence $$mathbb 1= (B-mathbb 1)(A-mathbb 1)=BA - A - B + mathbb 1$$



It follows that $$BA=A+B=AB$$ and we are done.



Note: here $mathbb 1$ denotes the appropriate identity matrix.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if $ a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ?
    $endgroup$
    – освящение
    Jan 19 '18 at 4:37












  • $begingroup$
    If yes, then please explain!
    $endgroup$
    – освящение
    Jan 19 '18 at 4:39






  • 1




    $begingroup$
    Yes, $X,Y$ are invertible matrices and these form a group.
    $endgroup$
    – lulu
    Jan 19 '18 at 10:47










  • $begingroup$
    but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$
    $endgroup$
    – освящение
    Jan 19 '18 at 15:10






  • 1




    $begingroup$
    @Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=mathbb 1$. If nothing else, that equation shows that $det X times det Y =1neq 0$ so both $X,Y$ have to be invertible,
    $endgroup$
    – lulu
    Jan 19 '18 at 15:14














34












34








34





$begingroup$

Consider the expression $$(A-mathbb 1)(B-mathbb 1)=AB-A-B+mathbb 1=mathbb 1$$



Thus $(A-mathbb 1)$ and $(B-mathbb 1)$ are inverse to each other, whence $$mathbb 1= (B-mathbb 1)(A-mathbb 1)=BA - A - B + mathbb 1$$



It follows that $$BA=A+B=AB$$ and we are done.



Note: here $mathbb 1$ denotes the appropriate identity matrix.






share|cite|improve this answer









$endgroup$



Consider the expression $$(A-mathbb 1)(B-mathbb 1)=AB-A-B+mathbb 1=mathbb 1$$



Thus $(A-mathbb 1)$ and $(B-mathbb 1)$ are inverse to each other, whence $$mathbb 1= (B-mathbb 1)(A-mathbb 1)=BA - A - B + mathbb 1$$



It follows that $$BA=A+B=AB$$ and we are done.



Note: here $mathbb 1$ denotes the appropriate identity matrix.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 19 '17 at 23:35









lulululu

42.4k25080




42.4k25080












  • $begingroup$
    if $ a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ?
    $endgroup$
    – освящение
    Jan 19 '18 at 4:37












  • $begingroup$
    If yes, then please explain!
    $endgroup$
    – освящение
    Jan 19 '18 at 4:39






  • 1




    $begingroup$
    Yes, $X,Y$ are invertible matrices and these form a group.
    $endgroup$
    – lulu
    Jan 19 '18 at 10:47










  • $begingroup$
    but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$
    $endgroup$
    – освящение
    Jan 19 '18 at 15:10






  • 1




    $begingroup$
    @Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=mathbb 1$. If nothing else, that equation shows that $det X times det Y =1neq 0$ so both $X,Y$ have to be invertible,
    $endgroup$
    – lulu
    Jan 19 '18 at 15:14


















  • $begingroup$
    if $ a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ?
    $endgroup$
    – освящение
    Jan 19 '18 at 4:37












  • $begingroup$
    If yes, then please explain!
    $endgroup$
    – освящение
    Jan 19 '18 at 4:39






  • 1




    $begingroup$
    Yes, $X,Y$ are invertible matrices and these form a group.
    $endgroup$
    – lulu
    Jan 19 '18 at 10:47










  • $begingroup$
    but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$
    $endgroup$
    – освящение
    Jan 19 '18 at 15:10






  • 1




    $begingroup$
    @Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=mathbb 1$. If nothing else, that equation shows that $det X times det Y =1neq 0$ so both $X,Y$ have to be invertible,
    $endgroup$
    – lulu
    Jan 19 '18 at 15:14
















$begingroup$
if $ a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ?
$endgroup$
– освящение
Jan 19 '18 at 4:37






$begingroup$
if $ a * b =e$ then $b*a=e$ iff $a$ and $b$ are elements of a group and $e$ is the identity element of that group . Are $X=(A-1)$ and $Y=(B-1)$elements of any group with identity $1$ ?
$endgroup$
– освящение
Jan 19 '18 at 4:37














$begingroup$
If yes, then please explain!
$endgroup$
– освящение
Jan 19 '18 at 4:39




$begingroup$
If yes, then please explain!
$endgroup$
– освящение
Jan 19 '18 at 4:39




1




1




$begingroup$
Yes, $X,Y$ are invertible matrices and these form a group.
$endgroup$
– lulu
Jan 19 '18 at 10:47




$begingroup$
Yes, $X,Y$ are invertible matrices and these form a group.
$endgroup$
– lulu
Jan 19 '18 at 10:47












$begingroup$
but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$
$endgroup$
– освящение
Jan 19 '18 at 15:10




$begingroup$
but how ? can you prove that $X$ and $Y$ are invertible with the condition $A+B=AB$
$endgroup$
– освящение
Jan 19 '18 at 15:10




1




1




$begingroup$
@Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=mathbb 1$. If nothing else, that equation shows that $det X times det Y =1neq 0$ so both $X,Y$ have to be invertible,
$endgroup$
– lulu
Jan 19 '18 at 15:14




$begingroup$
@Abhishek I proved they were invertible by exhibiting inverses. Specifically, I showed that $XY=mathbb 1$. If nothing else, that equation shows that $det X times det Y =1neq 0$ so both $X,Y$ have to be invertible,
$endgroup$
– lulu
Jan 19 '18 at 15:14











1












$begingroup$

Note: $I$ is Identity Matrix



Start with the equation given by the problem:
$$A+B=AB$$
left multiply both sides of this equation by B that produces:
$$B(A+B)=BAB$$
$$BA+BB=BAB$$
$$BA+BB=(BA)B$$
$$BB=(BA)B - BA$$
$$BB=(BA)(B - I);;(1)$$



then right multiply both sides of this equation by B that produces:



$$(A+B)B=ABB$$
$$AB+BB=ABB$$
$$AB+BB=(AB)B$$
$$BB=(AB)B - AB$$
$$BB=(AB)(B - I);;(2)$$



and we subtract equation(1) from equation(2) that gives:
$$0 = (AB - BA)(B - I)$$
if $B = I$,the equation $A+B=AB$ is broken. So, we can know $(AB - BA) = 0$ that means $$AB = BA$$



The purpose of this problem is to find the BA combination, we can get it by left multiplying.



Sorry,
if $B - I$ is a singular matrix, then $AB - BA$ may not be $0$.
So, I will try to find $M(B - I) = I$,
as above, start with equation:
$$A+B=AB$$
replace $B$ with $(B - I) + I$ in left side
$$A+(B - I) + I =AB$$
$$I =AB - A - (B - I)$$
$$I =A(B - I) - (B - I)$$
$$I =(A - I)(B - I)$$
the inverse matrix of $(B - I)$ is $(A - I)$, so $(B - I)$ is invertible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=begin{bmatrix}1&0\0&0end{bmatrix}$ and $Y=begin{bmatrix}0&0\0&1end{bmatrix}$
    $endgroup$
    – obareey
    Dec 26 '18 at 12:53












  • $begingroup$
    @obareey thanks, I have re-edited
    $endgroup$
    – Revc_Ra
    Dec 26 '18 at 15:59
















1












$begingroup$

Note: $I$ is Identity Matrix



Start with the equation given by the problem:
$$A+B=AB$$
left multiply both sides of this equation by B that produces:
$$B(A+B)=BAB$$
$$BA+BB=BAB$$
$$BA+BB=(BA)B$$
$$BB=(BA)B - BA$$
$$BB=(BA)(B - I);;(1)$$



then right multiply both sides of this equation by B that produces:



$$(A+B)B=ABB$$
$$AB+BB=ABB$$
$$AB+BB=(AB)B$$
$$BB=(AB)B - AB$$
$$BB=(AB)(B - I);;(2)$$



and we subtract equation(1) from equation(2) that gives:
$$0 = (AB - BA)(B - I)$$
if $B = I$,the equation $A+B=AB$ is broken. So, we can know $(AB - BA) = 0$ that means $$AB = BA$$



The purpose of this problem is to find the BA combination, we can get it by left multiplying.



Sorry,
if $B - I$ is a singular matrix, then $AB - BA$ may not be $0$.
So, I will try to find $M(B - I) = I$,
as above, start with equation:
$$A+B=AB$$
replace $B$ with $(B - I) + I$ in left side
$$A+(B - I) + I =AB$$
$$I =AB - A - (B - I)$$
$$I =A(B - I) - (B - I)$$
$$I =(A - I)(B - I)$$
the inverse matrix of $(B - I)$ is $(A - I)$, so $(B - I)$ is invertible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=begin{bmatrix}1&0\0&0end{bmatrix}$ and $Y=begin{bmatrix}0&0\0&1end{bmatrix}$
    $endgroup$
    – obareey
    Dec 26 '18 at 12:53












  • $begingroup$
    @obareey thanks, I have re-edited
    $endgroup$
    – Revc_Ra
    Dec 26 '18 at 15:59














1












1








1





$begingroup$

Note: $I$ is Identity Matrix



Start with the equation given by the problem:
$$A+B=AB$$
left multiply both sides of this equation by B that produces:
$$B(A+B)=BAB$$
$$BA+BB=BAB$$
$$BA+BB=(BA)B$$
$$BB=(BA)B - BA$$
$$BB=(BA)(B - I);;(1)$$



then right multiply both sides of this equation by B that produces:



$$(A+B)B=ABB$$
$$AB+BB=ABB$$
$$AB+BB=(AB)B$$
$$BB=(AB)B - AB$$
$$BB=(AB)(B - I);;(2)$$



and we subtract equation(1) from equation(2) that gives:
$$0 = (AB - BA)(B - I)$$
if $B = I$,the equation $A+B=AB$ is broken. So, we can know $(AB - BA) = 0$ that means $$AB = BA$$



The purpose of this problem is to find the BA combination, we can get it by left multiplying.



Sorry,
if $B - I$ is a singular matrix, then $AB - BA$ may not be $0$.
So, I will try to find $M(B - I) = I$,
as above, start with equation:
$$A+B=AB$$
replace $B$ with $(B - I) + I$ in left side
$$A+(B - I) + I =AB$$
$$I =AB - A - (B - I)$$
$$I =A(B - I) - (B - I)$$
$$I =(A - I)(B - I)$$
the inverse matrix of $(B - I)$ is $(A - I)$, so $(B - I)$ is invertible.






share|cite|improve this answer











$endgroup$



Note: $I$ is Identity Matrix



Start with the equation given by the problem:
$$A+B=AB$$
left multiply both sides of this equation by B that produces:
$$B(A+B)=BAB$$
$$BA+BB=BAB$$
$$BA+BB=(BA)B$$
$$BB=(BA)B - BA$$
$$BB=(BA)(B - I);;(1)$$



then right multiply both sides of this equation by B that produces:



$$(A+B)B=ABB$$
$$AB+BB=ABB$$
$$AB+BB=(AB)B$$
$$BB=(AB)B - AB$$
$$BB=(AB)(B - I);;(2)$$



and we subtract equation(1) from equation(2) that gives:
$$0 = (AB - BA)(B - I)$$
if $B = I$,the equation $A+B=AB$ is broken. So, we can know $(AB - BA) = 0$ that means $$AB = BA$$



The purpose of this problem is to find the BA combination, we can get it by left multiplying.



Sorry,
if $B - I$ is a singular matrix, then $AB - BA$ may not be $0$.
So, I will try to find $M(B - I) = I$,
as above, start with equation:
$$A+B=AB$$
replace $B$ with $(B - I) + I$ in left side
$$A+(B - I) + I =AB$$
$$I =AB - A - (B - I)$$
$$I =A(B - I) - (B - I)$$
$$I =(A - I)(B - I)$$
the inverse matrix of $(B - I)$ is $(A - I)$, so $(B - I)$ is invertible.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 15:28

























answered Dec 26 '18 at 11:20









Revc_RaRevc_Ra

112




112












  • $begingroup$
    $XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=begin{bmatrix}1&0\0&0end{bmatrix}$ and $Y=begin{bmatrix}0&0\0&1end{bmatrix}$
    $endgroup$
    – obareey
    Dec 26 '18 at 12:53












  • $begingroup$
    @obareey thanks, I have re-edited
    $endgroup$
    – Revc_Ra
    Dec 26 '18 at 15:59


















  • $begingroup$
    $XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=begin{bmatrix}1&0\0&0end{bmatrix}$ and $Y=begin{bmatrix}0&0\0&1end{bmatrix}$
    $endgroup$
    – obareey
    Dec 26 '18 at 12:53












  • $begingroup$
    @obareey thanks, I have re-edited
    $endgroup$
    – Revc_Ra
    Dec 26 '18 at 15:59
















$begingroup$
$XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=begin{bmatrix}1&0\0&0end{bmatrix}$ and $Y=begin{bmatrix}0&0\0&1end{bmatrix}$
$endgroup$
– obareey
Dec 26 '18 at 12:53






$begingroup$
$XY=0$ doesn't mean either $X$ or $Y$ is zero when $X$ and $Y$ are matrices. A counter example: $X=begin{bmatrix}1&0\0&0end{bmatrix}$ and $Y=begin{bmatrix}0&0\0&1end{bmatrix}$
$endgroup$
– obareey
Dec 26 '18 at 12:53














$begingroup$
@obareey thanks, I have re-edited
$endgroup$
– Revc_Ra
Dec 26 '18 at 15:59




$begingroup$
@obareey thanks, I have re-edited
$endgroup$
– Revc_Ra
Dec 26 '18 at 15:59


















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