If $A+B=AB$ then $AB=BA$
$begingroup$
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
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
$endgroup$
add a comment |
$begingroup$
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
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
$endgroup$
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
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
edited Dec 26 '18 at 11:24
Maria Mazur
46.7k1260120
46.7k1260120
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
asked Apr 19 '17 at 23:20
Sachchidanand PrasadSachchidanand Prasad
1,674722
1,674722
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 26 '18 at 11:20
Revc_RaRevc_Ra
112
$endgroup$
$begingroup$
$XY=0$ 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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2242629%2fif-ab-ab-then-ab-ba%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
$endgroup$
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.
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
answered Apr 19 '17 at 23:35
lulululu
42.9k25080
42.9k25080
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 26 '18 at 11:20
Revc_RaRevc_Ra
112
$endgroup$
$begingroup$
$XY=0$ 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
add a comment |
$begingroup$
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.
answered Dec 26 '18 at 11:20
Revc_RaRevc_Ra
112
$endgroup$
$begingroup$
$XY=0$ 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
add a comment |
$begingroup$
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.
answered Dec 26 '18 at 11:20
Revc_RaRevc_Ra
112
$endgroup$
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.
answered Dec 26 '18 at 11:20
Revc_RaRevc_Ra
112
edited Dec 26 '18 at 15:28
answered Dec 26 '18 at 11:20
Revc_RaRevc_Ra
112
answered Dec 26 '18 at 11:20
Revc_RaRevc_Ra
112
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
add a comment |
$begingroup$
$XY=0$ 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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2242629%2fif-ab-ab-then-ab-ba%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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