Alternative ways to determine the centre of $GL_n(K)$












1














Does anyone know a non calculative way the determine the centre of $GL_n(K)$?

Could we for example classify all normal subgroups of $GL_n(K)$ and unite them? (I know this will not work [look at $SL_n(K)$ for example, this normal subgroup is allready to big] it's just an example of the kind of proof I'm looking for).

I thought about this problem for some hours and I guess there just is no way to proof that the centre is ${lambda E , lambda in K^*}$ without "opening the group" and doing some matrix calculations.






linear-algebra abstract-algebra alternative-proof







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  • you can do it using the interpretation of $GL_n(k)= Aut(k^n)$ and using suitable morphisms (for example one that onnly permutes basis elements for a given basis), and then you see that it operates on every basis element the same and you are done. however, this correspnds precisely to the calculations you meant, but maybe it sheds more light on what is going on
    – Enkidu
    Dec 5 '18 at 12:33






  • 2




    Some of the answers to this question could be interesting.
    – Arnaud D.
    Dec 5 '18 at 12:35






  • 1




    You cannot avoid a calculation somewhere in the proof. Of course there are different approaches on different levels. But it doesn't seem to be very natural to give a proof on matrix groups without using matrix multiplication.
    – Dietrich Burde
    Dec 5 '18 at 14:09


















1














Does anyone know a non calculative way the determine the centre of $GL_n(K)$?

Could we for example classify all normal subgroups of $GL_n(K)$ and unite them? (I know this will not work [look at $SL_n(K)$ for example, this normal subgroup is allready to big] it's just an example of the kind of proof I'm looking for).

I thought about this problem for some hours and I guess there just is no way to proof that the centre is ${lambda E , lambda in K^*}$ without "opening the group" and doing some matrix calculations.






linear-algebra abstract-algebra alternative-proof







share|cite|improve this question
























  • you can do it using the interpretation of $GL_n(k)= Aut(k^n)$ and using suitable morphisms (for example one that onnly permutes basis elements for a given basis), and then you see that it operates on every basis element the same and you are done. however, this correspnds precisely to the calculations you meant, but maybe it sheds more light on what is going on
    – Enkidu
    Dec 5 '18 at 12:33






  • 2




    Some of the answers to this question could be interesting.
    – Arnaud D.
    Dec 5 '18 at 12:35






  • 1




    You cannot avoid a calculation somewhere in the proof. Of course there are different approaches on different levels. But it doesn't seem to be very natural to give a proof on matrix groups without using matrix multiplication.
    – Dietrich Burde
    Dec 5 '18 at 14:09
















1












1








1







Does anyone know a non calculative way the determine the centre of $GL_n(K)$?

Could we for example classify all normal subgroups of $GL_n(K)$ and unite them? (I know this will not work [look at $SL_n(K)$ for example, this normal subgroup is allready to big] it's just an example of the kind of proof I'm looking for).

I thought about this problem for some hours and I guess there just is no way to proof that the centre is ${lambda E , lambda in K^*}$ without "opening the group" and doing some matrix calculations.






linear-algebra abstract-algebra alternative-proof







share|cite|improve this question















Does anyone know a non calculative way the determine the centre of $GL_n(K)$?

Could we for example classify all normal subgroups of $GL_n(K)$ and unite them? (I know this will not work [look at $SL_n(K)$ for example, this normal subgroup is allready to big] it's just an example of the kind of proof I'm looking for).

I thought about this problem for some hours and I guess there just is no way to proof that the centre is ${lambda E , lambda in K^*}$ without "opening the group" and doing some matrix calculations.






linear-algebra abstract-algebra alternative-proof




linear-algebra abstract-algebra alternative-proof






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edited Dec 5 '18 at 12:33









Omnomnomnom

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asked Dec 5 '18 at 12:27









DominikDominik

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965












  • you can do it using the interpretation of $GL_n(k)= Aut(k^n)$ and using suitable morphisms (for example one that onnly permutes basis elements for a given basis), and then you see that it operates on every basis element the same and you are done. however, this correspnds precisely to the calculations you meant, but maybe it sheds more light on what is going on
    – Enkidu
    Dec 5 '18 at 12:33






  • 2




    Some of the answers to this question could be interesting.
    – Arnaud D.
    Dec 5 '18 at 12:35






  • 1




    You cannot avoid a calculation somewhere in the proof. Of course there are different approaches on different levels. But it doesn't seem to be very natural to give a proof on matrix groups without using matrix multiplication.
    – Dietrich Burde
    Dec 5 '18 at 14:09




















  • you can do it using the interpretation of $GL_n(k)= Aut(k^n)$ and using suitable morphisms (for example one that onnly permutes basis elements for a given basis), and then you see that it operates on every basis element the same and you are done. however, this correspnds precisely to the calculations you meant, but maybe it sheds more light on what is going on
    – Enkidu
    Dec 5 '18 at 12:33






  • 2




    Some of the answers to this question could be interesting.
    – Arnaud D.
    Dec 5 '18 at 12:35






  • 1




    You cannot avoid a calculation somewhere in the proof. Of course there are different approaches on different levels. But it doesn't seem to be very natural to give a proof on matrix groups without using matrix multiplication.
    – Dietrich Burde
    Dec 5 '18 at 14:09


















you can do it using the interpretation of $GL_n(k)= Aut(k^n)$ and using suitable morphisms (for example one that onnly permutes basis elements for a given basis), and then you see that it operates on every basis element the same and you are done. however, this correspnds precisely to the calculations you meant, but maybe it sheds more light on what is going on
– Enkidu
Dec 5 '18 at 12:33




you can do it using the interpretation of $GL_n(k)= Aut(k^n)$ and using suitable morphisms (for example one that onnly permutes basis elements for a given basis), and then you see that it operates on every basis element the same and you are done. however, this correspnds precisely to the calculations you meant, but maybe it sheds more light on what is going on
– Enkidu
Dec 5 '18 at 12:33




2




2




Some of the answers to this question could be interesting.
– Arnaud D.
Dec 5 '18 at 12:35




Some of the answers to this question could be interesting.
– Arnaud D.
Dec 5 '18 at 12:35




1




1




You cannot avoid a calculation somewhere in the proof. Of course there are different approaches on different levels. But it doesn't seem to be very natural to give a proof on matrix groups without using matrix multiplication.
– Dietrich Burde
Dec 5 '18 at 14:09






You cannot avoid a calculation somewhere in the proof. Of course there are different approaches on different levels. But it doesn't seem to be very natural to give a proof on matrix groups without using matrix multiplication.
– Dietrich Burde
Dec 5 '18 at 14:09












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First, you can show that the center of $M_n(k)$ is $k$ (embedded as the diagonal matrices) using the double centralizer theorem. This part is very conceptual. Next, you can show that any element in the center of $GL_n(k)$ must lie in the center of $M_n(k)$ by showing that every element of $M_n(k)$ is a linear combination of elements of $GL_n(k)$.



If $k$ is an infinite field this is easy: then if $X in M_n(k)$ is any matrix the polynomial $det (X + t I)$ is nonzero and hence takes nonzero values for all but finitely many nonzero values $t in k$. If $t_0 in k$ is nonzero and such that $det (X + tI) neq 0$, then $X$ is the difference of the invertible matrices $t I$ and $X + t I$.



I don't have quite so clean a way to handle the finite case. It suffices to write down a basis of $M_n(k)$ consisting of invertible matrices. For starters we can consider the identity matrix $I$ together with matrices of the form $I + E_{ij}$ where $E_{i,j}$ is the matrix with a $1$ in the $i,j$ entry and $0$ otherwise, and also $i neq j$. Next, to handle the diagonal, if the characteristic of $k$ isn't equal to $2$ we can consider the diagonal matrices with entries $(1, -1, 1, dots), (1, 1, -1, dots)$, etc. If the characteristic is $2$ we can consider matrices of the form $I - E_{i,i} + E_{(i-1),i} + E_{i,i-1}$ for $i = 2, dots n$.






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    First, you can show that the center of $M_n(k)$ is $k$ (embedded as the diagonal matrices) using the double centralizer theorem. This part is very conceptual. Next, you can show that any element in the center of $GL_n(k)$ must lie in the center of $M_n(k)$ by showing that every element of $M_n(k)$ is a linear combination of elements of $GL_n(k)$.



    If $k$ is an infinite field this is easy: then if $X in M_n(k)$ is any matrix the polynomial $det (X + t I)$ is nonzero and hence takes nonzero values for all but finitely many nonzero values $t in k$. If $t_0 in k$ is nonzero and such that $det (X + tI) neq 0$, then $X$ is the difference of the invertible matrices $t I$ and $X + t I$.



    I don't have quite so clean a way to handle the finite case. It suffices to write down a basis of $M_n(k)$ consisting of invertible matrices. For starters we can consider the identity matrix $I$ together with matrices of the form $I + E_{ij}$ where $E_{i,j}$ is the matrix with a $1$ in the $i,j$ entry and $0$ otherwise, and also $i neq j$. Next, to handle the diagonal, if the characteristic of $k$ isn't equal to $2$ we can consider the diagonal matrices with entries $(1, -1, 1, dots), (1, 1, -1, dots)$, etc. If the characteristic is $2$ we can consider matrices of the form $I - E_{i,i} + E_{(i-1),i} + E_{i,i-1}$ for $i = 2, dots n$.






    share|cite|improve this answer


























      1














      First, you can show that the center of $M_n(k)$ is $k$ (embedded as the diagonal matrices) using the double centralizer theorem. This part is very conceptual. Next, you can show that any element in the center of $GL_n(k)$ must lie in the center of $M_n(k)$ by showing that every element of $M_n(k)$ is a linear combination of elements of $GL_n(k)$.



      If $k$ is an infinite field this is easy: then if $X in M_n(k)$ is any matrix the polynomial $det (X + t I)$ is nonzero and hence takes nonzero values for all but finitely many nonzero values $t in k$. If $t_0 in k$ is nonzero and such that $det (X + tI) neq 0$, then $X$ is the difference of the invertible matrices $t I$ and $X + t I$.



      I don't have quite so clean a way to handle the finite case. It suffices to write down a basis of $M_n(k)$ consisting of invertible matrices. For starters we can consider the identity matrix $I$ together with matrices of the form $I + E_{ij}$ where $E_{i,j}$ is the matrix with a $1$ in the $i,j$ entry and $0$ otherwise, and also $i neq j$. Next, to handle the diagonal, if the characteristic of $k$ isn't equal to $2$ we can consider the diagonal matrices with entries $(1, -1, 1, dots), (1, 1, -1, dots)$, etc. If the characteristic is $2$ we can consider matrices of the form $I - E_{i,i} + E_{(i-1),i} + E_{i,i-1}$ for $i = 2, dots n$.






      share|cite|improve this answer
























        1












        1








        1






        First, you can show that the center of $M_n(k)$ is $k$ (embedded as the diagonal matrices) using the double centralizer theorem. This part is very conceptual. Next, you can show that any element in the center of $GL_n(k)$ must lie in the center of $M_n(k)$ by showing that every element of $M_n(k)$ is a linear combination of elements of $GL_n(k)$.



        If $k$ is an infinite field this is easy: then if $X in M_n(k)$ is any matrix the polynomial $det (X + t I)$ is nonzero and hence takes nonzero values for all but finitely many nonzero values $t in k$. If $t_0 in k$ is nonzero and such that $det (X + tI) neq 0$, then $X$ is the difference of the invertible matrices $t I$ and $X + t I$.



        I don't have quite so clean a way to handle the finite case. It suffices to write down a basis of $M_n(k)$ consisting of invertible matrices. For starters we can consider the identity matrix $I$ together with matrices of the form $I + E_{ij}$ where $E_{i,j}$ is the matrix with a $1$ in the $i,j$ entry and $0$ otherwise, and also $i neq j$. Next, to handle the diagonal, if the characteristic of $k$ isn't equal to $2$ we can consider the diagonal matrices with entries $(1, -1, 1, dots), (1, 1, -1, dots)$, etc. If the characteristic is $2$ we can consider matrices of the form $I - E_{i,i} + E_{(i-1),i} + E_{i,i-1}$ for $i = 2, dots n$.






        share|cite|improve this answer












        First, you can show that the center of $M_n(k)$ is $k$ (embedded as the diagonal matrices) using the double centralizer theorem. This part is very conceptual. Next, you can show that any element in the center of $GL_n(k)$ must lie in the center of $M_n(k)$ by showing that every element of $M_n(k)$ is a linear combination of elements of $GL_n(k)$.



        If $k$ is an infinite field this is easy: then if $X in M_n(k)$ is any matrix the polynomial $det (X + t I)$ is nonzero and hence takes nonzero values for all but finitely many nonzero values $t in k$. If $t_0 in k$ is nonzero and such that $det (X + tI) neq 0$, then $X$ is the difference of the invertible matrices $t I$ and $X + t I$.



        I don't have quite so clean a way to handle the finite case. It suffices to write down a basis of $M_n(k)$ consisting of invertible matrices. For starters we can consider the identity matrix $I$ together with matrices of the form $I + E_{ij}$ where $E_{i,j}$ is the matrix with a $1$ in the $i,j$ entry and $0$ otherwise, and also $i neq j$. Next, to handle the diagonal, if the characteristic of $k$ isn't equal to $2$ we can consider the diagonal matrices with entries $(1, -1, 1, dots), (1, 1, -1, dots)$, etc. If the characteristic is $2$ we can consider matrices of the form $I - E_{i,i} + E_{(i-1),i} + E_{i,i-1}$ for $i = 2, dots n$.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Dec 5 '18 at 22:54









        Qiaochu YuanQiaochu Yuan

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