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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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
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
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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
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
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
add a comment |
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
add a comment |
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 $$A^lB^k=B^kA^l$$
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
add a comment |
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.
Thank you for your explanation.
– Carl Schmidt
Nov 22 at 21:47
add a comment |
Your Answer
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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 $$A^lB^k=B^kA^l$$
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
add a comment |
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 $$A^lB^k=B^kA^l$$
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
add a comment |
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 $$A^lB^k=B^kA^l$$
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 $$A^lB^k=B^kA^l$$
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
add a comment |
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
add a comment |
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.
Thank you for your explanation.
– Carl Schmidt
Nov 22 at 21:47
add a comment |
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.
Thank you for your explanation.
– Carl Schmidt
Nov 22 at 21:47
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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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