Transvection matrices generate $ operatorname{SL}_n(mathbb{R}) $
$begingroup$
I need to prove that the transvection matrices generate the special linear group $operatorname{SL}_n left(mathbb{R}right) $.
I want to proceed using induction on $n$.
I was able to prove the $2times 2$ case, but I am having difficulty with the $n+1$ case.
I supposed that the elementary matrices of the first type generate $operatorname{SL}_n(mathbb{R})$. And I want to show that an elementary matrix of the first type of order $n+1$ can generate $operatorname{SL}_{n+1}(mathbb{R})$
matrices group-theory algebraic-k-theory
edited Jan 23 at 12:46
amWhy
1
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
$endgroup$
add a comment |
$begingroup$
I need to prove that the transvection matrices generate the special linear group $operatorname{SL}_n left(mathbb{R}right) $.
I want to proceed using induction on $n$.
I was able to prove the $2times 2$ case, but I am having difficulty with the $n+1$ case.
I supposed that the elementary matrices of the first type generate $operatorname{SL}_n(mathbb{R})$. And I want to show that an elementary matrix of the first type of order $n+1$ can generate $operatorname{SL}_{n+1}(mathbb{R})$
matrices group-theory algebraic-k-theory
edited Jan 23 at 12:46
amWhy
1
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
$endgroup$
2
$begingroup$
Possible duplicate of What are the generators for $SL_n(mathbb{R})$ (Michael Artin's Algebra book)
$endgroup$
– user549397
Jan 24 at 3:25
add a comment |
$begingroup$
I need to prove that the transvection matrices generate the special linear group $operatorname{SL}_n left(mathbb{R}right) $.
I want to proceed using induction on $n$.
I was able to prove the $2times 2$ case, but I am having difficulty with the $n+1$ case.
I supposed that the elementary matrices of the first type generate $operatorname{SL}_n(mathbb{R})$. And I want to show that an elementary matrix of the first type of order $n+1$ can generate $operatorname{SL}_{n+1}(mathbb{R})$
matrices group-theory algebraic-k-theory
edited Jan 23 at 12:46
amWhy
1
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
$endgroup$
I need to prove that the transvection matrices generate the special linear group $operatorname{SL}_n left(mathbb{R}right) $.
I want to proceed using induction on $n$.
I was able to prove the $2times 2$ case, but I am having difficulty with the $n+1$ case.
I supposed that the elementary matrices of the first type generate $operatorname{SL}_n(mathbb{R})$. And I want to show that an elementary matrix of the first type of order $n+1$ can generate $operatorname{SL}_{n+1}(mathbb{R})$
matrices group-theory algebraic-k-theory
matrices group-theory algebraic-k-theory
edited Jan 23 at 12:46
amWhy
1
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
edited Jan 23 at 12:46
amWhy
1
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
edited Jan 23 at 12:46
amWhy
1
edited Jan 23 at 12:46
amWhy
1
edited Jan 23 at 12:46
amWhy
1
1
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
asked Sep 6 '13 at 10:30
CarpediemCarpediem
832632
832632
2
$begingroup$
Possible duplicate of What are the generators for $SL_n(mathbb{R})$ (Michael Artin's Algebra book)
$endgroup$
– user549397
Jan 24 at 3:25
add a comment |
2
$begingroup$
Possible duplicate of What are the generators for $SL_n(mathbb{R})$ (Michael Artin's Algebra book)
$endgroup$
– user549397
Jan 24 at 3:25
2
2
$begingroup$
Possible duplicate of What are the generators for $SL_n(mathbb{R})$ (Michael Artin's Algebra book)
$endgroup$
– user549397
Jan 24 at 3:25
$begingroup$
Possible duplicate of What are the generators for $SL_n(mathbb{R})$ (Michael Artin's Algebra book)
$endgroup$
– user549397
Jan 24 at 3:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$.
The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) in {rm SL}_n(K)$.
If $a_{12} = 0$ then, choose some $j$ with $a_{1j} ne 0$ and add column $j$ to column $2$ to get $a_{12} ne 0$.
Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.
For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$. Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
You have now reduced to the case $n-1$.
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
$endgroup$
add a comment |
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$begingroup$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$.
The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) in {rm SL}_n(K)$.
If $a_{12} = 0$ then, choose some $j$ with $a_{1j} ne 0$ and add column $j$ to column $2$ to get $a_{12} ne 0$.
Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.
For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$. Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
You have now reduced to the case $n-1$.
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
$endgroup$
add a comment |
$begingroup$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$.
The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) in {rm SL}_n(K)$.
If $a_{12} = 0$ then, choose some $j$ with $a_{1j} ne 0$ and add column $j$ to column $2$ to get $a_{12} ne 0$.
Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.
For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$. Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
You have now reduced to the case $n-1$.
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
$endgroup$
add a comment |
$begingroup$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$.
The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) in {rm SL}_n(K)$.
If $a_{12} = 0$ then, choose some $j$ with $a_{1j} ne 0$ and add column $j$ to column $2$ to get $a_{12} ne 0$.
Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.
For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$. Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
You have now reduced to the case $n-1$.
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
$endgroup$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$.
The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) in {rm SL}_n(K)$.
If $a_{12} = 0$ then, choose some $j$ with $a_{1j} ne 0$ and add column $j$ to column $2$ to get $a_{12} ne 0$.
Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.
For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$. Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
You have now reduced to the case $n-1$.
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
edited Nov 29 '16 at 9:47
darij grinberg
11.1k33167
11.1k33167
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
answered Sep 6 '13 at 10:49
Derek HoltDerek Holt
53.8k53571
53.8k53571
add a comment |
add a comment |
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$begingroup$
Possible duplicate of What are the generators for $SL_n(mathbb{R})$ (Michael Artin's Algebra book)
$endgroup$
– user549397
Jan 24 at 3:25