Show that, for square matrices $A$ and $B$, $A+B=AB$ implies $AB=BA$.

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Let $A$ and $B$ be two $n$-by-$n$ real matrices such that $A+B = AB$.
How do I prove that $AB= BA$?




I have tried using the trace function on $A+B-AB$. But I could not get any Ideas.
Kindly provide me with hints.






linear-algebra matrices







share|cite|improve this question



















  • 1




    I want to remark that, for a vector space $V$ over a field $K$ and for two $K$-linear operators $A,B: Vto V$ such that $A+B=Acirc B$, if $V$ is finite-dimensional over $K$, then $A$ and $B$ commute (i.e., $Acirc B=Bcirc A$). However, the result does not hold if $V$ is infinite-dimensional over $K$.
    – Batominovski
    6 mins ago







  • 1




    Take, for example, $V$ to be the vector space of infinite sequences $mathbfx:=left(x_iright)_iinmathbbZ_>0$ of elements $x_1,x_2,x_3,ldotsin K$. For each $mathbfx:=left(x_iright)_iinmathbbZ_>0in V$, define $$A(mathbfx):=left(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_5,x_5-x_6,ldotsright)$$ and $$B(mathbfx):=left(x_1,x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,ldotsright),.$$
    – Batominovski
    6 mins ago







  • 1




    Then, for every $mathbfx:=(x_i)_iinmathbbZ_>0$ in $V$, $$(A+B)(mathbfx)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright)$$ and $$(Acirc B)(mathbfx)=Aleft(x_1,x_2-x_1,x_3-x_2,ldotsright)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),,$$ but $$(Bcirc A)(mathbfx)=Bleft(x_1-x_2,x_2-x_3,x_3-x_4,ldotsright)=left(x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),.$$ Thus, $$A+B=Acirc Btext but Acirc Bneq Bcirc A,.$$
    – Batominovski
    5 mins ago















up vote
2
down vote

favorite













Let $A$ and $B$ be two $n$-by-$n$ real matrices such that $A+B = AB$.
How do I prove that $AB= BA$?




I have tried using the trace function on $A+B-AB$. But I could not get any Ideas.
Kindly provide me with hints.






linear-algebra matrices







share|cite|improve this question



















  • 1




    I want to remark that, for a vector space $V$ over a field $K$ and for two $K$-linear operators $A,B: Vto V$ such that $A+B=Acirc B$, if $V$ is finite-dimensional over $K$, then $A$ and $B$ commute (i.e., $Acirc B=Bcirc A$). However, the result does not hold if $V$ is infinite-dimensional over $K$.
    – Batominovski
    6 mins ago







  • 1




    Take, for example, $V$ to be the vector space of infinite sequences $mathbfx:=left(x_iright)_iinmathbbZ_>0$ of elements $x_1,x_2,x_3,ldotsin K$. For each $mathbfx:=left(x_iright)_iinmathbbZ_>0in V$, define $$A(mathbfx):=left(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_5,x_5-x_6,ldotsright)$$ and $$B(mathbfx):=left(x_1,x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,ldotsright),.$$
    – Batominovski
    6 mins ago







  • 1




    Then, for every $mathbfx:=(x_i)_iinmathbbZ_>0$ in $V$, $$(A+B)(mathbfx)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright)$$ and $$(Acirc B)(mathbfx)=Aleft(x_1,x_2-x_1,x_3-x_2,ldotsright)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),,$$ but $$(Bcirc A)(mathbfx)=Bleft(x_1-x_2,x_2-x_3,x_3-x_4,ldotsright)=left(x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),.$$ Thus, $$A+B=Acirc Btext but Acirc Bneq Bcirc A,.$$
    – Batominovski
    5 mins ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite












Let $A$ and $B$ be two $n$-by-$n$ real matrices such that $A+B = AB$.
How do I prove that $AB= BA$?




I have tried using the trace function on $A+B-AB$. But I could not get any Ideas.
Kindly provide me with hints.






linear-algebra matrices







share|cite|improve this question
















Let $A$ and $B$ be two $n$-by-$n$ real matrices such that $A+B = AB$.
How do I prove that $AB= BA$?




I have tried using the trace function on $A+B-AB$. But I could not get any Ideas.
Kindly provide me with hints.






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 49 secs ago









Batominovski

27.5k22881




27.5k22881










asked 24 mins ago









tony

1489




1489







  • 1




    I want to remark that, for a vector space $V$ over a field $K$ and for two $K$-linear operators $A,B: Vto V$ such that $A+B=Acirc B$, if $V$ is finite-dimensional over $K$, then $A$ and $B$ commute (i.e., $Acirc B=Bcirc A$). However, the result does not hold if $V$ is infinite-dimensional over $K$.
    – Batominovski
    6 mins ago







  • 1




    Take, for example, $V$ to be the vector space of infinite sequences $mathbfx:=left(x_iright)_iinmathbbZ_>0$ of elements $x_1,x_2,x_3,ldotsin K$. For each $mathbfx:=left(x_iright)_iinmathbbZ_>0in V$, define $$A(mathbfx):=left(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_5,x_5-x_6,ldotsright)$$ and $$B(mathbfx):=left(x_1,x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,ldotsright),.$$
    – Batominovski
    6 mins ago







  • 1




    Then, for every $mathbfx:=(x_i)_iinmathbbZ_>0$ in $V$, $$(A+B)(mathbfx)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright)$$ and $$(Acirc B)(mathbfx)=Aleft(x_1,x_2-x_1,x_3-x_2,ldotsright)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),,$$ but $$(Bcirc A)(mathbfx)=Bleft(x_1-x_2,x_2-x_3,x_3-x_4,ldotsright)=left(x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),.$$ Thus, $$A+B=Acirc Btext but Acirc Bneq Bcirc A,.$$
    – Batominovski
    5 mins ago













  • 1




    I want to remark that, for a vector space $V$ over a field $K$ and for two $K$-linear operators $A,B: Vto V$ such that $A+B=Acirc B$, if $V$ is finite-dimensional over $K$, then $A$ and $B$ commute (i.e., $Acirc B=Bcirc A$). However, the result does not hold if $V$ is infinite-dimensional over $K$.
    – Batominovski
    6 mins ago







  • 1




    Take, for example, $V$ to be the vector space of infinite sequences $mathbfx:=left(x_iright)_iinmathbbZ_>0$ of elements $x_1,x_2,x_3,ldotsin K$. For each $mathbfx:=left(x_iright)_iinmathbbZ_>0in V$, define $$A(mathbfx):=left(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_5,x_5-x_6,ldotsright)$$ and $$B(mathbfx):=left(x_1,x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,ldotsright),.$$
    – Batominovski
    6 mins ago







  • 1




    Then, for every $mathbfx:=(x_i)_iinmathbbZ_>0$ in $V$, $$(A+B)(mathbfx)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright)$$ and $$(Acirc B)(mathbfx)=Aleft(x_1,x_2-x_1,x_3-x_2,ldotsright)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),,$$ but $$(Bcirc A)(mathbfx)=Bleft(x_1-x_2,x_2-x_3,x_3-x_4,ldotsright)=left(x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),.$$ Thus, $$A+B=Acirc Btext but Acirc Bneq Bcirc A,.$$
    – Batominovski
    5 mins ago








1




1




I want to remark that, for a vector space $V$ over a field $K$ and for two $K$-linear operators $A,B: Vto V$ such that $A+B=Acirc B$, if $V$ is finite-dimensional over $K$, then $A$ and $B$ commute (i.e., $Acirc B=Bcirc A$). However, the result does not hold if $V$ is infinite-dimensional over $K$.
– Batominovski
6 mins ago





I want to remark that, for a vector space $V$ over a field $K$ and for two $K$-linear operators $A,B: Vto V$ such that $A+B=Acirc B$, if $V$ is finite-dimensional over $K$, then $A$ and $B$ commute (i.e., $Acirc B=Bcirc A$). However, the result does not hold if $V$ is infinite-dimensional over $K$.
– Batominovski
6 mins ago





1




1




Take, for example, $V$ to be the vector space of infinite sequences $mathbfx:=left(x_iright)_iinmathbbZ_>0$ of elements $x_1,x_2,x_3,ldotsin K$. For each $mathbfx:=left(x_iright)_iinmathbbZ_>0in V$, define $$A(mathbfx):=left(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_5,x_5-x_6,ldotsright)$$ and $$B(mathbfx):=left(x_1,x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,ldotsright),.$$
– Batominovski
6 mins ago





Take, for example, $V$ to be the vector space of infinite sequences $mathbfx:=left(x_iright)_iinmathbbZ_>0$ of elements $x_1,x_2,x_3,ldotsin K$. For each $mathbfx:=left(x_iright)_iinmathbbZ_>0in V$, define $$A(mathbfx):=left(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_5,x_5-x_6,ldotsright)$$ and $$B(mathbfx):=left(x_1,x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,ldotsright),.$$
– Batominovski
6 mins ago





1




1




Then, for every $mathbfx:=(x_i)_iinmathbbZ_>0$ in $V$, $$(A+B)(mathbfx)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright)$$ and $$(Acirc B)(mathbfx)=Aleft(x_1,x_2-x_1,x_3-x_2,ldotsright)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),,$$ but $$(Bcirc A)(mathbfx)=Bleft(x_1-x_2,x_2-x_3,x_3-x_4,ldotsright)=left(x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),.$$ Thus, $$A+B=Acirc Btext but Acirc Bneq Bcirc A,.$$
– Batominovski
5 mins ago





Then, for every $mathbfx:=(x_i)_iinmathbbZ_>0$ in $V$, $$(A+B)(mathbfx)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright)$$ and $$(Acirc B)(mathbfx)=Aleft(x_1,x_2-x_1,x_3-x_2,ldotsright)=left(2x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),,$$ but $$(Bcirc A)(mathbfx)=Bleft(x_1-x_2,x_2-x_3,x_3-x_4,ldotsright)=left(x_1-x_2,2x_2-x_1-x_3,2x_3-x_2-x_4,ldotsright),.$$ Thus, $$A+B=Acirc Btext but Acirc Bneq Bcirc A,.$$
– Batominovski
5 mins ago











2 Answers
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$A+B=AB$ is equivalent to $(I-A)(I-B)=I$. As $I-A$ and $I-B$ are square, this implies $(I-B)(I-A)=I$, etc.






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  • @tony then mark it as answered.
    – user25959
    14 mins ago










  • how do i do that?
    – tony
    13 mins ago






  • 1




    @user25959 You cannot accept an answer before 15 min. after the post.
    – cansomeonehelpmeout
    12 mins ago

















up vote
-3
down vote













Addition of matrices is always commutative, so $$BA = B + A = A + B = AB$$






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  • 5




    How do you know that $BA = B + A$?
    – Theo Bendit
    18 mins ago










  • ...I don't think that this deserves the down vote, as the commutative addition $A + B = B + A$ implies that swapping the values does not change the result.
    – simon.watts
    1 min ago










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2 Answers
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active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
7
down vote













$A+B=AB$ is equivalent to $(I-A)(I-B)=I$. As $I-A$ and $I-B$ are square, this implies $(I-B)(I-A)=I$, etc.






share|cite|improve this answer




















  • @tony then mark it as answered.
    – user25959
    14 mins ago










  • how do i do that?
    – tony
    13 mins ago






  • 1




    @user25959 You cannot accept an answer before 15 min. after the post.
    – cansomeonehelpmeout
    12 mins ago














up vote
7
down vote













$A+B=AB$ is equivalent to $(I-A)(I-B)=I$. As $I-A$ and $I-B$ are square, this implies $(I-B)(I-A)=I$, etc.






share|cite|improve this answer




















  • @tony then mark it as answered.
    – user25959
    14 mins ago










  • how do i do that?
    – tony
    13 mins ago






  • 1




    @user25959 You cannot accept an answer before 15 min. after the post.
    – cansomeonehelpmeout
    12 mins ago












up vote
7
down vote










up vote
7
down vote









$A+B=AB$ is equivalent to $(I-A)(I-B)=I$. As $I-A$ and $I-B$ are square, this implies $(I-B)(I-A)=I$, etc.






share|cite|improve this answer












$A+B=AB$ is equivalent to $(I-A)(I-B)=I$. As $I-A$ and $I-B$ are square, this implies $(I-B)(I-A)=I$, etc.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 22 mins ago









Lord Shark the Unknown

92k955118




92k955118











  • @tony then mark it as answered.
    – user25959
    14 mins ago










  • how do i do that?
    – tony
    13 mins ago






  • 1




    @user25959 You cannot accept an answer before 15 min. after the post.
    – cansomeonehelpmeout
    12 mins ago
















  • @tony then mark it as answered.
    – user25959
    14 mins ago










  • how do i do that?
    – tony
    13 mins ago






  • 1




    @user25959 You cannot accept an answer before 15 min. after the post.
    – cansomeonehelpmeout
    12 mins ago















@tony then mark it as answered.
– user25959
14 mins ago




@tony then mark it as answered.
– user25959
14 mins ago












how do i do that?
– tony
13 mins ago




how do i do that?
– tony
13 mins ago




1




1




@user25959 You cannot accept an answer before 15 min. after the post.
– cansomeonehelpmeout
12 mins ago




@user25959 You cannot accept an answer before 15 min. after the post.
– cansomeonehelpmeout
12 mins ago










up vote
-3
down vote













Addition of matrices is always commutative, so $$BA = B + A = A + B = AB$$






share|cite|improve this answer
















  • 5




    How do you know that $BA = B + A$?
    – Theo Bendit
    18 mins ago










  • ...I don't think that this deserves the down vote, as the commutative addition $A + B = B + A$ implies that swapping the values does not change the result.
    – simon.watts
    1 min ago














up vote
-3
down vote













Addition of matrices is always commutative, so $$BA = B + A = A + B = AB$$






share|cite|improve this answer
















  • 5




    How do you know that $BA = B + A$?
    – Theo Bendit
    18 mins ago










  • ...I don't think that this deserves the down vote, as the commutative addition $A + B = B + A$ implies that swapping the values does not change the result.
    – simon.watts
    1 min ago












up vote
-3
down vote










up vote
-3
down vote









Addition of matrices is always commutative, so $$BA = B + A = A + B = AB$$






share|cite|improve this answer












Addition of matrices is always commutative, so $$BA = B + A = A + B = AB$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 19 mins ago









Lukas Kofler

7931518




7931518







  • 5




    How do you know that $BA = B + A$?
    – Theo Bendit
    18 mins ago










  • ...I don't think that this deserves the down vote, as the commutative addition $A + B = B + A$ implies that swapping the values does not change the result.
    – simon.watts
    1 min ago












  • 5




    How do you know that $BA = B + A$?
    – Theo Bendit
    18 mins ago










  • ...I don't think that this deserves the down vote, as the commutative addition $A + B = B + A$ implies that swapping the values does not change the result.
    – simon.watts
    1 min ago







5




5




How do you know that $BA = B + A$?
– Theo Bendit
18 mins ago




How do you know that $BA = B + A$?
– Theo Bendit
18 mins ago












...I don't think that this deserves the down vote, as the commutative addition $A + B = B + A$ implies that swapping the values does not change the result.
– simon.watts
1 min ago




...I don't think that this deserves the down vote, as the commutative addition $A + B = B + A$ implies that swapping the values does not change the result.
– simon.watts
1 min ago

















 

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