Given $A in SL_n(Bbb R)$ Find $B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$
$begingroup$
I was trying to prove that, $[SL_n(Bbb R), SL_n(Bbb R)]=SL_n(Bbb R)$
My attempt:
$[SL_n(Bbb R), SL_n(Bbb R)]subset SL_n(Bbb R)$ is trivial, trying to prove the other inclusion.
So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ .
In fact, I proved that $SL_n(Bbb R)$ is generated by elementary matrices of the form ${I_n+lambda e_{ij} | lambdain Bbb R , i ne j}$.
So the problem can be reduced to :
Given an elementary matrix of the above form say, $E=I_n+lambda e_{ij}$ find , $B,P in SL_n(Bbb R)$ such that $E=BPB^{-1}P^{-1}$
Thanks in advance for help!
linear-algebra linear-groups
asked Jan 15 at 16:00
CoherentCoherent
1,176523
$endgroup$
add a comment |
$begingroup$
I was trying to prove that, $[SL_n(Bbb R), SL_n(Bbb R)]=SL_n(Bbb R)$
My attempt:
$[SL_n(Bbb R), SL_n(Bbb R)]subset SL_n(Bbb R)$ is trivial, trying to prove the other inclusion.
So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ .
In fact, I proved that $SL_n(Bbb R)$ is generated by elementary matrices of the form ${I_n+lambda e_{ij} | lambdain Bbb R , i ne j}$.
So the problem can be reduced to :
Given an elementary matrix of the above form say, $E=I_n+lambda e_{ij}$ find , $B,P in SL_n(Bbb R)$ such that $E=BPB^{-1}P^{-1}$
Thanks in advance for help!
linear-algebra linear-groups
asked Jan 15 at 16:00
CoherentCoherent
1,176523
$endgroup$
$begingroup$
Hint: Can you show that the only normal subgroups of $operatorname{SL}_{n}(mathbb{R})$ are $left{ operatorname{Id} right}, left{ pm operatorname{Id} right}$ (when $n$ is even), and $operatorname{SL}_{n}(mathbb{R})$ itself? Then since $left[ operatorname{SL}_{n}(mathbb{R}), operatorname{SL}_{n}(mathbb{R}) right]$ is a normal subgroup, all you need rule out is that it is not $left{ pm operatorname{Id} right} $ when $n$ is even.
$endgroup$
– Adam Higgins
Jan 15 at 16:10
$begingroup$
"So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ ." Yes, that's enough to show. But I'm far from sure that it is correct. Being a product of commutators is a weaker property than being a commutator oneself.
$endgroup$
– darij grinberg
Jan 15 at 17:06
add a comment |
$begingroup$
I was trying to prove that, $[SL_n(Bbb R), SL_n(Bbb R)]=SL_n(Bbb R)$
My attempt:
$[SL_n(Bbb R), SL_n(Bbb R)]subset SL_n(Bbb R)$ is trivial, trying to prove the other inclusion.
So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ .
In fact, I proved that $SL_n(Bbb R)$ is generated by elementary matrices of the form ${I_n+lambda e_{ij} | lambdain Bbb R , i ne j}$.
So the problem can be reduced to :
Given an elementary matrix of the above form say, $E=I_n+lambda e_{ij}$ find , $B,P in SL_n(Bbb R)$ such that $E=BPB^{-1}P^{-1}$
Thanks in advance for help!
linear-algebra linear-groups
asked Jan 15 at 16:00
CoherentCoherent
1,176523
$endgroup$
I was trying to prove that, $[SL_n(Bbb R), SL_n(Bbb R)]=SL_n(Bbb R)$
My attempt:
$[SL_n(Bbb R), SL_n(Bbb R)]subset SL_n(Bbb R)$ is trivial, trying to prove the other inclusion.
So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ .
In fact, I proved that $SL_n(Bbb R)$ is generated by elementary matrices of the form ${I_n+lambda e_{ij} | lambdain Bbb R , i ne j}$.
So the problem can be reduced to :
Given an elementary matrix of the above form say, $E=I_n+lambda e_{ij}$ find , $B,P in SL_n(Bbb R)$ such that $E=BPB^{-1}P^{-1}$
Thanks in advance for help!
linear-algebra linear-groups
linear-algebra linear-groups
asked Jan 15 at 16:00
CoherentCoherent
1,176523
asked Jan 15 at 16:00
CoherentCoherent
1,176523
asked Jan 15 at 16:00
CoherentCoherent
1,176523
asked Jan 15 at 16:00
CoherentCoherent
1,176523
asked Jan 15 at 16:00
CoherentCoherent
1,176523
1,176523
$begingroup$
Hint: Can you show that the only normal subgroups of $operatorname{SL}_{n}(mathbb{R})$ are $left{ operatorname{Id} right}, left{ pm operatorname{Id} right}$ (when $n$ is even), and $operatorname{SL}_{n}(mathbb{R})$ itself? Then since $left[ operatorname{SL}_{n}(mathbb{R}), operatorname{SL}_{n}(mathbb{R}) right]$ is a normal subgroup, all you need rule out is that it is not $left{ pm operatorname{Id} right} $ when $n$ is even.
$endgroup$
– Adam Higgins
Jan 15 at 16:10
$begingroup$
"So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ ." Yes, that's enough to show. But I'm far from sure that it is correct. Being a product of commutators is a weaker property than being a commutator oneself.
$endgroup$
– darij grinberg
Jan 15 at 17:06
add a comment |
$begingroup$
Hint: Can you show that the only normal subgroups of $operatorname{SL}_{n}(mathbb{R})$ are $left{ operatorname{Id} right}, left{ pm operatorname{Id} right}$ (when $n$ is even), and $operatorname{SL}_{n}(mathbb{R})$ itself? Then since $left[ operatorname{SL}_{n}(mathbb{R}), operatorname{SL}_{n}(mathbb{R}) right]$ is a normal subgroup, all you need rule out is that it is not $left{ pm operatorname{Id} right} $ when $n$ is even.
$endgroup$
– Adam Higgins
Jan 15 at 16:10
$begingroup$
"So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ ." Yes, that's enough to show. But I'm far from sure that it is correct. Being a product of commutators is a weaker property than being a commutator oneself.
$endgroup$
– darij grinberg
Jan 15 at 17:06
$begingroup$
Hint: Can you show that the only normal subgroups of $operatorname{SL}_{n}(mathbb{R})$ are $left{ operatorname{Id} right}, left{ pm operatorname{Id} right}$ (when $n$ is even), and $operatorname{SL}_{n}(mathbb{R})$ itself? Then since $left[ operatorname{SL}_{n}(mathbb{R}), operatorname{SL}_{n}(mathbb{R}) right]$ is a normal subgroup, all you need rule out is that it is not $left{ pm operatorname{Id} right} $ when $n$ is even.
$endgroup$
– Adam Higgins
Jan 15 at 16:10
$begingroup$
Hint: Can you show that the only normal subgroups of $operatorname{SL}_{n}(mathbb{R})$ are $left{ operatorname{Id} right}, left{ pm operatorname{Id} right}$ (when $n$ is even), and $operatorname{SL}_{n}(mathbb{R})$ itself? Then since $left[ operatorname{SL}_{n}(mathbb{R}), operatorname{SL}_{n}(mathbb{R}) right]$ is a normal subgroup, all you need rule out is that it is not $left{ pm operatorname{Id} right} $ when $n$ is even.
$endgroup$
– Adam Higgins
Jan 15 at 16:10
$begingroup$
"So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ ." Yes, that's enough to show. But I'm far from sure that it is correct. Being a product of commutators is a weaker property than being a commutator oneself.
$endgroup$
– darij grinberg
Jan 15 at 17:06
$begingroup$
"So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ ." Yes, that's enough to show. But I'm far from sure that it is correct. Being a product of commutators is a weaker property than being a commutator oneself.
$endgroup$
– darij grinberg
Jan 15 at 17:06
add a comment |
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$begingroup$
Hint: Can you show that the only normal subgroups of $operatorname{SL}_{n}(mathbb{R})$ are $left{ operatorname{Id} right}, left{ pm operatorname{Id} right}$ (when $n$ is even), and $operatorname{SL}_{n}(mathbb{R})$ itself? Then since $left[ operatorname{SL}_{n}(mathbb{R}), operatorname{SL}_{n}(mathbb{R}) right]$ is a normal subgroup, all you need rule out is that it is not $left{ pm operatorname{Id} right} $ when $n$ is even.
$endgroup$
– Adam Higgins
Jan 15 at 16:10
$begingroup$
"So it's enough to show that given any $A in SL_n(Bbb R)$ $exists B,P in SL_n(Bbb R)$ such that, $A=BPB^{-1}P^{-1}$ ." Yes, that's enough to show. But I'm far from sure that it is correct. Being a product of commutators is a weaker property than being a commutator oneself.
$endgroup$
– darij grinberg
Jan 15 at 17:06