Matrix units in commutator subgroup
$begingroup$
Let $m ge 2$ be an integer, $R = M_{m}(K)$ be the algebra of all $mtimes m$ matrices over $K$ an infinite field, if $L = [R,R]$ and $a,p in R$ are non-zero elements. Then, if $[au, pu] = 0$ for all $u in L$, then $a= alpha p$ for some $alpha in K$.
I am reading a proof of above lemma and what I don't get is the following:
If $a = Sigma_{ij} a_{ij}e_{ij}$ and $p =Sigma_{ij} p_{ij}e_{ij}$ for some $a_{ij},p_{ij} in Kbackslash {0}$ with $1 le i,j le m$. Then, for every $1 le i neq j le m$ we have $0 = e_{kk}[ae_{ij},pe_{ij}] = (a_{ki}p_{ji} - p_{ki}a_{ji})e_{kj}$
I can prove the first equality but have trouble with the second one. Is there some trick to see this or is this just plain calculation? I have tried to calculate the bracket explicitly by writing terms, but it gets a more complicated cause of matrix unit elements ($e_{ij}$). Maybe I missed some trivial detail, any suggestions would be much appreciated. Thanks.
linear-algebra matrices ring-theory
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
$endgroup$
add a comment |
$begingroup$
Let $m ge 2$ be an integer, $R = M_{m}(K)$ be the algebra of all $mtimes m$ matrices over $K$ an infinite field, if $L = [R,R]$ and $a,p in R$ are non-zero elements. Then, if $[au, pu] = 0$ for all $u in L$, then $a= alpha p$ for some $alpha in K$.
I am reading a proof of above lemma and what I don't get is the following:
If $a = Sigma_{ij} a_{ij}e_{ij}$ and $p =Sigma_{ij} p_{ij}e_{ij}$ for some $a_{ij},p_{ij} in Kbackslash {0}$ with $1 le i,j le m$. Then, for every $1 le i neq j le m$ we have $0 = e_{kk}[ae_{ij},pe_{ij}] = (a_{ki}p_{ji} - p_{ki}a_{ji})e_{kj}$
I can prove the first equality but have trouble with the second one. Is there some trick to see this or is this just plain calculation? I have tried to calculate the bracket explicitly by writing terms, but it gets a more complicated cause of matrix unit elements ($e_{ij}$). Maybe I missed some trivial detail, any suggestions would be much appreciated. Thanks.
linear-algebra matrices ring-theory
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
$endgroup$
add a comment |
$begingroup$
Let $m ge 2$ be an integer, $R = M_{m}(K)$ be the algebra of all $mtimes m$ matrices over $K$ an infinite field, if $L = [R,R]$ and $a,p in R$ are non-zero elements. Then, if $[au, pu] = 0$ for all $u in L$, then $a= alpha p$ for some $alpha in K$.
I am reading a proof of above lemma and what I don't get is the following:
If $a = Sigma_{ij} a_{ij}e_{ij}$ and $p =Sigma_{ij} p_{ij}e_{ij}$ for some $a_{ij},p_{ij} in Kbackslash {0}$ with $1 le i,j le m$. Then, for every $1 le i neq j le m$ we have $0 = e_{kk}[ae_{ij},pe_{ij}] = (a_{ki}p_{ji} - p_{ki}a_{ji})e_{kj}$
I can prove the first equality but have trouble with the second one. Is there some trick to see this or is this just plain calculation? I have tried to calculate the bracket explicitly by writing terms, but it gets a more complicated cause of matrix unit elements ($e_{ij}$). Maybe I missed some trivial detail, any suggestions would be much appreciated. Thanks.
linear-algebra matrices ring-theory
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
$endgroup$
Let $m ge 2$ be an integer, $R = M_{m}(K)$ be the algebra of all $mtimes m$ matrices over $K$ an infinite field, if $L = [R,R]$ and $a,p in R$ are non-zero elements. Then, if $[au, pu] = 0$ for all $u in L$, then $a= alpha p$ for some $alpha in K$.
I am reading a proof of above lemma and what I don't get is the following:
If $a = Sigma_{ij} a_{ij}e_{ij}$ and $p =Sigma_{ij} p_{ij}e_{ij}$ for some $a_{ij},p_{ij} in Kbackslash {0}$ with $1 le i,j le m$. Then, for every $1 le i neq j le m$ we have $0 = e_{kk}[ae_{ij},pe_{ij}] = (a_{ki}p_{ji} - p_{ki}a_{ji})e_{kj}$
I can prove the first equality but have trouble with the second one. Is there some trick to see this or is this just plain calculation? I have tried to calculate the bracket explicitly by writing terms, but it gets a more complicated cause of matrix unit elements ($e_{ij}$). Maybe I missed some trivial detail, any suggestions would be much appreciated. Thanks.
linear-algebra matrices ring-theory
linear-algebra matrices ring-theory
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
edited Jan 17 at 16:44
Mayank Mishra
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
asked Jan 17 at 13:10
Mayank MishraMayank Mishra
1318
1318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It can be obtained by direct calculation:
begin{aligned}
e_{kk}[ae_{ij},pe_{ij}]
&=e_{kk}ae_{ij}pe_{ij}-e_{kk}pe_{ij}ae_{ij}\
&=e_k(e_k^Tae_i)(e_j^Tpe_i)e_j^T-e_k(e_k^Tpe_i)(e_j^Tae_i)e_j^T\
&=e_k(a_{ki}p_{ji}-p_{ki}a_{ji})e_j^T\
&=(a_{ki}p_{ji}-p_{ki}a_{ji})e_{kj}.
end{aligned}
answered Jan 17 at 14:56
user1551user1551
74.3k566129
$endgroup$
$begingroup$
I got the trick in step 1 but I didn't get how you go from step 2 to step 3?
$endgroup$
– Mayank Mishra
Jan 17 at 16:54
$begingroup$
@Rick In general, $e_k^TM$ is the $k$-th row of a matrix $M$ and $Me_i$ is the $i$-th column of $M$. Therefore, $e_k^Tae_i$ is the $i$-th column of the $k$-th row of $a$, i.e. the $(k,i)$-th entry of $a$, or $a_{ki}$.
$endgroup$
– user1551
Jan 17 at 17:03
$begingroup$
Sorry for my unfamiliarity with matrix units, but what I read from wiki, isn't $e_{k}^{T}M = 0$ as they are product of $1times k$ and $mtimes m$ matrix?
$endgroup$
– Mayank Mishra
Jan 17 at 17:11
$begingroup$
@Rick ${e_1,e_2,ldots,e_m}$ denotes the canonical basis of $K^m$ here. That is, $e_kin K^m$ is a column vector with $m$ entries, with a $1$ at the $k$-th position and zeroes elsewhere. E.g. when $m=3$, we have $e_1=(1,0,0)^T,,e_2=(0,1,0)^T$ and $e_3=(0,0,1)^T$. This is almost a standard notation in linear algebra. With this definition, we have $e_{rs}=e_re_s^T$ for any $r,s$.
$endgroup$
– user1551
Jan 17 at 17:43
$begingroup$
Oh, it was simple, just got confused with notations. Thanks.
$endgroup$
– Mayank Mishra
Jan 17 at 17:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It can be obtained by direct calculation:
begin{aligned}
e_{kk}[ae_{ij},pe_{ij}]
&=e_{kk}ae_{ij}pe_{ij}-e_{kk}pe_{ij}ae_{ij}\
&=e_k(e_k^Tae_i)(e_j^Tpe_i)e_j^T-e_k(e_k^Tpe_i)(e_j^Tae_i)e_j^T\
&=e_k(a_{ki}p_{ji}-p_{ki}a_{ji})e_j^T\
&=(a_{ki}p_{ji}-p_{ki}a_{ji})e_{kj}.
end{aligned}
answered Jan 17 at 14:56
user1551user1551
74.3k566129
$endgroup$
$begingroup$
I got the trick in step 1 but I didn't get how you go from step 2 to step 3?
$endgroup$
– Mayank Mishra
Jan 17 at 16:54
$begingroup$
@Rick In general, $e_k^TM$ is the $k$-th row of a matrix $M$ and $Me_i$ is the $i$-th column of $M$. Therefore, $e_k^Tae_i$ is the $i$-th column of the $k$-th row of $a$, i.e. the $(k,i)$-th entry of $a$, or $a_{ki}$.
$endgroup$
– user1551
Jan 17 at 17:03
$begingroup$
Sorry for my unfamiliarity with matrix units, but what I read from wiki, isn't $e_{k}^{T}M = 0$ as they are product of $1times k$ and $mtimes m$ matrix?
$endgroup$
– Mayank Mishra
Jan 17 at 17:11
$begingroup$
@Rick ${e_1,e_2,ldots,e_m}$ denotes the canonical basis of $K^m$ here. That is, $e_kin K^m$ is a column vector with $m$ entries, with a $1$ at the $k$-th position and zeroes elsewhere. E.g. when $m=3$, we have $e_1=(1,0,0)^T,,e_2=(0,1,0)^T$ and $e_3=(0,0,1)^T$. This is almost a standard notation in linear algebra. With this definition, we have $e_{rs}=e_re_s^T$ for any $r,s$.
$endgroup$
– user1551
Jan 17 at 17:43
$begingroup$
Oh, it was simple, just got confused with notations. Thanks.
$endgroup$
– Mayank Mishra
Jan 17 at 17:49
add a comment |
$begingroup$
It can be obtained by direct calculation:
begin{aligned}
e_{kk}[ae_{ij},pe_{ij}]
&=e_{kk}ae_{ij}pe_{ij}-e_{kk}pe_{ij}ae_{ij}\
&=e_k(e_k^Tae_i)(e_j^Tpe_i)e_j^T-e_k(e_k^Tpe_i)(e_j^Tae_i)e_j^T\
&=e_k(a_{ki}p_{ji}-p_{ki}a_{ji})e_j^T\
&=(a_{ki}p_{ji}-p_{ki}a_{ji})e_{kj}.
end{aligned}
answered Jan 17 at 14:56
user1551user1551
74.3k566129
$endgroup$
$begingroup$
I got the trick in step 1 but I didn't get how you go from step 2 to step 3?
$endgroup$
– Mayank Mishra
Jan 17 at 16:54
$begingroup$
@Rick In general, $e_k^TM$ is the $k$-th row of a matrix $M$ and $Me_i$ is the $i$-th column of $M$. Therefore, $e_k^Tae_i$ is the $i$-th column of the $k$-th row of $a$, i.e. the $(k,i)$-th entry of $a$, or $a_{ki}$.
$endgroup$
– user1551
Jan 17 at 17:03
$begingroup$
Sorry for my unfamiliarity with matrix units, but what I read from wiki, isn't $e_{k}^{T}M = 0$ as they are product of $1times k$ and $mtimes m$ matrix?
$endgroup$
– Mayank Mishra
Jan 17 at 17:11
$begingroup$
@Rick ${e_1,e_2,ldots,e_m}$ denotes the canonical basis of $K^m$ here. That is, $e_kin K^m$ is a column vector with $m$ entries, with a $1$ at the $k$-th position and zeroes elsewhere. E.g. when $m=3$, we have $e_1=(1,0,0)^T,,e_2=(0,1,0)^T$ and $e_3=(0,0,1)^T$. This is almost a standard notation in linear algebra. With this definition, we have $e_{rs}=e_re_s^T$ for any $r,s$.
$endgroup$
– user1551
Jan 17 at 17:43
$begingroup$
Oh, it was simple, just got confused with notations. Thanks.
$endgroup$
– Mayank Mishra
Jan 17 at 17:49
add a comment |
$begingroup$
It can be obtained by direct calculation:
begin{aligned}
e_{kk}[ae_{ij},pe_{ij}]
&=e_{kk}ae_{ij}pe_{ij}-e_{kk}pe_{ij}ae_{ij}\
&=e_k(e_k^Tae_i)(e_j^Tpe_i)e_j^T-e_k(e_k^Tpe_i)(e_j^Tae_i)e_j^T\
&=e_k(a_{ki}p_{ji}-p_{ki}a_{ji})e_j^T\
&=(a_{ki}p_{ji}-p_{ki}a_{ji})e_{kj}.
end{aligned}
answered Jan 17 at 14:56
user1551user1551
74.3k566129
$endgroup$
It can be obtained by direct calculation:
begin{aligned}
e_{kk}[ae_{ij},pe_{ij}]
&=e_{kk}ae_{ij}pe_{ij}-e_{kk}pe_{ij}ae_{ij}\
&=e_k(e_k^Tae_i)(e_j^Tpe_i)e_j^T-e_k(e_k^Tpe_i)(e_j^Tae_i)e_j^T\
&=e_k(a_{ki}p_{ji}-p_{ki}a_{ji})e_j^T\
&=(a_{ki}p_{ji}-p_{ki}a_{ji})e_{kj}.
end{aligned}
answered Jan 17 at 14:56
user1551user1551
74.3k566129
answered Jan 17 at 14:56
user1551user1551
74.3k566129
answered Jan 17 at 14:56
user1551user1551
74.3k566129
answered Jan 17 at 14:56
user1551user1551
74.3k566129
74.3k566129
$begingroup$
I got the trick in step 1 but I didn't get how you go from step 2 to step 3?
$endgroup$
– Mayank Mishra
Jan 17 at 16:54
$begingroup$
@Rick In general, $e_k^TM$ is the $k$-th row of a matrix $M$ and $Me_i$ is the $i$-th column of $M$. Therefore, $e_k^Tae_i$ is the $i$-th column of the $k$-th row of $a$, i.e. the $(k,i)$-th entry of $a$, or $a_{ki}$.
$endgroup$
– user1551
Jan 17 at 17:03
$begingroup$
Sorry for my unfamiliarity with matrix units, but what I read from wiki, isn't $e_{k}^{T}M = 0$ as they are product of $1times k$ and $mtimes m$ matrix?
$endgroup$
– Mayank Mishra
Jan 17 at 17:11
$begingroup$
@Rick ${e_1,e_2,ldots,e_m}$ denotes the canonical basis of $K^m$ here. That is, $e_kin K^m$ is a column vector with $m$ entries, with a $1$ at the $k$-th position and zeroes elsewhere. E.g. when $m=3$, we have $e_1=(1,0,0)^T,,e_2=(0,1,0)^T$ and $e_3=(0,0,1)^T$. This is almost a standard notation in linear algebra. With this definition, we have $e_{rs}=e_re_s^T$ for any $r,s$.
$endgroup$
– user1551
Jan 17 at 17:43
$begingroup$
Oh, it was simple, just got confused with notations. Thanks.
$endgroup$
– Mayank Mishra
Jan 17 at 17:49
add a comment |
$begingroup$
I got the trick in step 1 but I didn't get how you go from step 2 to step 3?
$endgroup$
– Mayank Mishra
Jan 17 at 16:54
$begingroup$
@Rick In general, $e_k^TM$ is the $k$-th row of a matrix $M$ and $Me_i$ is the $i$-th column of $M$. Therefore, $e_k^Tae_i$ is the $i$-th column of the $k$-th row of $a$, i.e. the $(k,i)$-th entry of $a$, or $a_{ki}$.
$endgroup$
– user1551
Jan 17 at 17:03
$begingroup$
Sorry for my unfamiliarity with matrix units, but what I read from wiki, isn't $e_{k}^{T}M = 0$ as they are product of $1times k$ and $mtimes m$ matrix?
$endgroup$
– Mayank Mishra
Jan 17 at 17:11
$begingroup$
@Rick ${e_1,e_2,ldots,e_m}$ denotes the canonical basis of $K^m$ here. That is, $e_kin K^m$ is a column vector with $m$ entries, with a $1$ at the $k$-th position and zeroes elsewhere. E.g. when $m=3$, we have $e_1=(1,0,0)^T,,e_2=(0,1,0)^T$ and $e_3=(0,0,1)^T$. This is almost a standard notation in linear algebra. With this definition, we have $e_{rs}=e_re_s^T$ for any $r,s$.
$endgroup$
– user1551
Jan 17 at 17:43
$begingroup$
Oh, it was simple, just got confused with notations. Thanks.
$endgroup$
– Mayank Mishra
Jan 17 at 17:49
$begingroup$
I got the trick in step 1 but I didn't get how you go from step 2 to step 3?
$endgroup$
– Mayank Mishra
Jan 17 at 16:54
$begingroup$
I got the trick in step 1 but I didn't get how you go from step 2 to step 3?
$endgroup$
– Mayank Mishra
Jan 17 at 16:54
$begingroup$
@Rick In general, $e_k^TM$ is the $k$-th row of a matrix $M$ and $Me_i$ is the $i$-th column of $M$. Therefore, $e_k^Tae_i$ is the $i$-th column of the $k$-th row of $a$, i.e. the $(k,i)$-th entry of $a$, or $a_{ki}$.
$endgroup$
– user1551
Jan 17 at 17:03
$begingroup$
@Rick In general, $e_k^TM$ is the $k$-th row of a matrix $M$ and $Me_i$ is the $i$-th column of $M$. Therefore, $e_k^Tae_i$ is the $i$-th column of the $k$-th row of $a$, i.e. the $(k,i)$-th entry of $a$, or $a_{ki}$.
$endgroup$
– user1551
Jan 17 at 17:03
$begingroup$
Sorry for my unfamiliarity with matrix units, but what I read from wiki, isn't $e_{k}^{T}M = 0$ as they are product of $1times k$ and $mtimes m$ matrix?
$endgroup$
– Mayank Mishra
Jan 17 at 17:11
$begingroup$
Sorry for my unfamiliarity with matrix units, but what I read from wiki, isn't $e_{k}^{T}M = 0$ as they are product of $1times k$ and $mtimes m$ matrix?
$endgroup$
– Mayank Mishra
Jan 17 at 17:11
$begingroup$
@Rick ${e_1,e_2,ldots,e_m}$ denotes the canonical basis of $K^m$ here. That is, $e_kin K^m$ is a column vector with $m$ entries, with a $1$ at the $k$-th position and zeroes elsewhere. E.g. when $m=3$, we have $e_1=(1,0,0)^T,,e_2=(0,1,0)^T$ and $e_3=(0,0,1)^T$. This is almost a standard notation in linear algebra. With this definition, we have $e_{rs}=e_re_s^T$ for any $r,s$.
$endgroup$
– user1551
Jan 17 at 17:43
$begingroup$
@Rick ${e_1,e_2,ldots,e_m}$ denotes the canonical basis of $K^m$ here. That is, $e_kin K^m$ is a column vector with $m$ entries, with a $1$ at the $k$-th position and zeroes elsewhere. E.g. when $m=3$, we have $e_1=(1,0,0)^T,,e_2=(0,1,0)^T$ and $e_3=(0,0,1)^T$. This is almost a standard notation in linear algebra. With this definition, we have $e_{rs}=e_re_s^T$ for any $r,s$.
$endgroup$
– user1551
Jan 17 at 17:43
$begingroup$
Oh, it was simple, just got confused with notations. Thanks.
$endgroup$
– Mayank Mishra
Jan 17 at 17:49
$begingroup$
Oh, it was simple, just got confused with notations. Thanks.
$endgroup$
– Mayank Mishra
Jan 17 at 17:49
add a comment |
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