Commutating Matrices, a question











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RE-EDITED:



I've come to conclusion that $e^{At}e^{Bt}=e^{Bt}e^{At}$ if:



$sum_{k=0}^{infty}sum_{j=0}^{infty}frac{t^{k+j}}{k!j!}A^kB^j=sum_{j=0}^{infty}sum_{k=0}^{infty}frac{t^{k+j}}{k!j!}B^jA^k$



Does anyone have any idea how I can verify that it only is true if $AB=BA$?






linear-algebra







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  • Yes, just apply $AB=BA$ repeatedly in $A^xB^y$, you will get $B^yA^x$. An example $A^2B^2=A(AB)B=A(BA)B=(AB)(AB)=(BA)(BA)=B(AB)A=B(BA)A=B^2A^2.$
    – John_Wick
    Nov 22 at 20:30















up vote
0
down vote

favorite












RE-EDITED:



I've come to conclusion that $e^{At}e^{Bt}=e^{Bt}e^{At}$ if:



$sum_{k=0}^{infty}sum_{j=0}^{infty}frac{t^{k+j}}{k!j!}A^kB^j=sum_{j=0}^{infty}sum_{k=0}^{infty}frac{t^{k+j}}{k!j!}B^jA^k$



Does anyone have any idea how I can verify that it only is true if $AB=BA$?






linear-algebra







share|cite|improve this question






















  • Yes, just apply $AB=BA$ repeatedly in $A^xB^y$, you will get $B^yA^x$. An example $A^2B^2=A(AB)B=A(BA)B=(AB)(AB)=(BA)(BA)=B(AB)A=B(BA)A=B^2A^2.$
    – John_Wick
    Nov 22 at 20:30













up vote
0
down vote

favorite









up vote
0
down vote

favorite











RE-EDITED:



I've come to conclusion that $e^{At}e^{Bt}=e^{Bt}e^{At}$ if:



$sum_{k=0}^{infty}sum_{j=0}^{infty}frac{t^{k+j}}{k!j!}A^kB^j=sum_{j=0}^{infty}sum_{k=0}^{infty}frac{t^{k+j}}{k!j!}B^jA^k$



Does anyone have any idea how I can verify that it only is true if $AB=BA$?






linear-algebra







share|cite|improve this question













RE-EDITED:



I've come to conclusion that $e^{At}e^{Bt}=e^{Bt}e^{At}$ if:



$sum_{k=0}^{infty}sum_{j=0}^{infty}frac{t^{k+j}}{k!j!}A^kB^j=sum_{j=0}^{infty}sum_{k=0}^{infty}frac{t^{k+j}}{k!j!}B^jA^k$



Does anyone have any idea how I can verify that it only is true if $AB=BA$?






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 20:24









Carl Schmidt

81




81












  • Yes, just apply $AB=BA$ repeatedly in $A^xB^y$, you will get $B^yA^x$. An example $A^2B^2=A(AB)B=A(BA)B=(AB)(AB)=(BA)(BA)=B(AB)A=B(BA)A=B^2A^2.$
    – John_Wick
    Nov 22 at 20:30


















  • Yes, just apply $AB=BA$ repeatedly in $A^xB^y$, you will get $B^yA^x$. An example $A^2B^2=A(AB)B=A(BA)B=(AB)(AB)=(BA)(BA)=B(AB)A=B(BA)A=B^2A^2.$
    – John_Wick
    Nov 22 at 20:30
















Yes, just apply $AB=BA$ repeatedly in $A^xB^y$, you will get $B^yA^x$. An example $A^2B^2=A(AB)B=A(BA)B=(AB)(AB)=(BA)(BA)=B(AB)A=B(BA)A=B^2A^2.$
– John_Wick
Nov 22 at 20:30




Yes, just apply $AB=BA$ repeatedly in $A^xB^y$, you will get $B^yA^x$. An example $A^2B^2=A(AB)B=A(BA)B=(AB)(AB)=(BA)(BA)=B(AB)A=B(BA)A=B^2A^2.$
– John_Wick
Nov 22 at 20:30










2 Answers
2






active

oldest

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0
down vote



accepted










Using induction we have $AB=BA$. Also if $A^mB^n=B^nA^m$ we obtain$$A^{m+1}B^n=A.A^mB^n=AB^nA^m=B^nA^{m+1}$$similarly$$A^mB^{n+1}=B^{n+1}A^m$$and $$A^{m+1}B^{n+1}=B^{n+1}A^{m+1}$$which means that for all natural $l,k$ we have $$A^lB^k=B^kA^l$$






share|cite|improve this answer





















  • Thank you for your explanation. This made it possible to finish a proof I've been working on for the last 5 hours...
    – Carl Schmidt
    Nov 22 at 21:48












  • I'm happy for you. Wish you luck!!!
    – Mostafa Ayaz
    Nov 22 at 21:49


















up vote
1
down vote













We have
begin{align}
A^x B^y &= underbrace{Acdots (A}_{x text{ times}}underbrace{B)cdots B}_{y text{ times}}\
&= underbrace{Acdots (B}_{x text{ times}}underbrace{A)cdots B}_{y text{ times}}\
&= underbrace{Acdots (A}_{x-1 text{ times}}B)Aunderbrace{Bcdots B}_{y-1 text{ times}}\
&= underbrace{Acdots (A}_{x-2 text{ times}}B)AAunderbrace{Bcdots B}_{y-1 text{ times}}\
&= ldots\
&=Bunderbrace{Acdots A}_{x text{ times}}underbrace{Bcdots B}_{y-1 text{ times}}\ &= BA^xB^{y-1}.
end{align}
Now repeat this $y-1$ more times.






share|cite|improve this answer























  • Thank you for your explanation.
    – Carl Schmidt
    Nov 22 at 21:47











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Using induction we have $AB=BA$. Also if $A^mB^n=B^nA^m$ we obtain$$A^{m+1}B^n=A.A^mB^n=AB^nA^m=B^nA^{m+1}$$similarly$$A^mB^{n+1}=B^{n+1}A^m$$and $$A^{m+1}B^{n+1}=B^{n+1}A^{m+1}$$which means that for all natural $l,k$ we have $$A^lB^k=B^kA^l$$






share|cite|improve this answer





















  • Thank you for your explanation. This made it possible to finish a proof I've been working on for the last 5 hours...
    – Carl Schmidt
    Nov 22 at 21:48












  • I'm happy for you. Wish you luck!!!
    – Mostafa Ayaz
    Nov 22 at 21:49















up vote
0
down vote



accepted










Using induction we have $AB=BA$. Also if $A^mB^n=B^nA^m$ we obtain$$A^{m+1}B^n=A.A^mB^n=AB^nA^m=B^nA^{m+1}$$similarly$$A^mB^{n+1}=B^{n+1}A^m$$and $$A^{m+1}B^{n+1}=B^{n+1}A^{m+1}$$which means that for all natural $l,k$ we have $$A^lB^k=B^kA^l$$






share|cite|improve this answer





















  • Thank you for your explanation. This made it possible to finish a proof I've been working on for the last 5 hours...
    – Carl Schmidt
    Nov 22 at 21:48












  • I'm happy for you. Wish you luck!!!
    – Mostafa Ayaz
    Nov 22 at 21:49













up vote
0
down vote



accepted







up vote
0
down vote



accepted






Using induction we have $AB=BA$. Also if $A^mB^n=B^nA^m$ we obtain$$A^{m+1}B^n=A.A^mB^n=AB^nA^m=B^nA^{m+1}$$similarly$$A^mB^{n+1}=B^{n+1}A^m$$and $$A^{m+1}B^{n+1}=B^{n+1}A^{m+1}$$which means that for all natural $l,k$ we have $$A^lB^k=B^kA^l$$






share|cite|improve this answer












Using induction we have $AB=BA$. Also if $A^mB^n=B^nA^m$ we obtain$$A^{m+1}B^n=A.A^mB^n=AB^nA^m=B^nA^{m+1}$$similarly$$A^mB^{n+1}=B^{n+1}A^m$$and $$A^{m+1}B^{n+1}=B^{n+1}A^{m+1}$$which means that for all natural $l,k$ we have $$A^lB^k=B^kA^l$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 21:16









Mostafa Ayaz

13.6k3836




13.6k3836












  • Thank you for your explanation. This made it possible to finish a proof I've been working on for the last 5 hours...
    – Carl Schmidt
    Nov 22 at 21:48












  • I'm happy for you. Wish you luck!!!
    – Mostafa Ayaz
    Nov 22 at 21:49


















  • Thank you for your explanation. This made it possible to finish a proof I've been working on for the last 5 hours...
    – Carl Schmidt
    Nov 22 at 21:48












  • I'm happy for you. Wish you luck!!!
    – Mostafa Ayaz
    Nov 22 at 21:49
















Thank you for your explanation. This made it possible to finish a proof I've been working on for the last 5 hours...
– Carl Schmidt
Nov 22 at 21:48






Thank you for your explanation. This made it possible to finish a proof I've been working on for the last 5 hours...
– Carl Schmidt
Nov 22 at 21:48














I'm happy for you. Wish you luck!!!
– Mostafa Ayaz
Nov 22 at 21:49




I'm happy for you. Wish you luck!!!
– Mostafa Ayaz
Nov 22 at 21:49










up vote
1
down vote













We have
begin{align}
A^x B^y &= underbrace{Acdots (A}_{x text{ times}}underbrace{B)cdots B}_{y text{ times}}\
&= underbrace{Acdots (B}_{x text{ times}}underbrace{A)cdots B}_{y text{ times}}\
&= underbrace{Acdots (A}_{x-1 text{ times}}B)Aunderbrace{Bcdots B}_{y-1 text{ times}}\
&= underbrace{Acdots (A}_{x-2 text{ times}}B)AAunderbrace{Bcdots B}_{y-1 text{ times}}\
&= ldots\
&=Bunderbrace{Acdots A}_{x text{ times}}underbrace{Bcdots B}_{y-1 text{ times}}\ &= BA^xB^{y-1}.
end{align}
Now repeat this $y-1$ more times.






share|cite|improve this answer























  • Thank you for your explanation.
    – Carl Schmidt
    Nov 22 at 21:47















up vote
1
down vote













We have
begin{align}
A^x B^y &= underbrace{Acdots (A}_{x text{ times}}underbrace{B)cdots B}_{y text{ times}}\
&= underbrace{Acdots (B}_{x text{ times}}underbrace{A)cdots B}_{y text{ times}}\
&= underbrace{Acdots (A}_{x-1 text{ times}}B)Aunderbrace{Bcdots B}_{y-1 text{ times}}\
&= underbrace{Acdots (A}_{x-2 text{ times}}B)AAunderbrace{Bcdots B}_{y-1 text{ times}}\
&= ldots\
&=Bunderbrace{Acdots A}_{x text{ times}}underbrace{Bcdots B}_{y-1 text{ times}}\ &= BA^xB^{y-1}.
end{align}
Now repeat this $y-1$ more times.






share|cite|improve this answer























  • Thank you for your explanation.
    – Carl Schmidt
    Nov 22 at 21:47













up vote
1
down vote










up vote
1
down vote









We have
begin{align}
A^x B^y &= underbrace{Acdots (A}_{x text{ times}}underbrace{B)cdots B}_{y text{ times}}\
&= underbrace{Acdots (B}_{x text{ times}}underbrace{A)cdots B}_{y text{ times}}\
&= underbrace{Acdots (A}_{x-1 text{ times}}B)Aunderbrace{Bcdots B}_{y-1 text{ times}}\
&= underbrace{Acdots (A}_{x-2 text{ times}}B)AAunderbrace{Bcdots B}_{y-1 text{ times}}\
&= ldots\
&=Bunderbrace{Acdots A}_{x text{ times}}underbrace{Bcdots B}_{y-1 text{ times}}\ &= BA^xB^{y-1}.
end{align}
Now repeat this $y-1$ more times.






share|cite|improve this answer














We have
begin{align}
A^x B^y &= underbrace{Acdots (A}_{x text{ times}}underbrace{B)cdots B}_{y text{ times}}\
&= underbrace{Acdots (B}_{x text{ times}}underbrace{A)cdots B}_{y text{ times}}\
&= underbrace{Acdots (A}_{x-1 text{ times}}B)Aunderbrace{Bcdots B}_{y-1 text{ times}}\
&= underbrace{Acdots (A}_{x-2 text{ times}}B)AAunderbrace{Bcdots B}_{y-1 text{ times}}\
&= ldots\
&=Bunderbrace{Acdots A}_{x text{ times}}underbrace{Bcdots B}_{y-1 text{ times}}\ &= BA^xB^{y-1}.
end{align}
Now repeat this $y-1$ more times.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 at 20:38

























answered Nov 22 at 20:31









MisterRiemann

5,7131624




5,7131624












  • Thank you for your explanation.
    – Carl Schmidt
    Nov 22 at 21:47


















  • Thank you for your explanation.
    – Carl Schmidt
    Nov 22 at 21:47
















Thank you for your explanation.
– Carl Schmidt
Nov 22 at 21:47




Thank you for your explanation.
– Carl Schmidt
Nov 22 at 21:47


















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