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Suppose $g$ is a symmetric bilinear form, is $operatornamerad g = operatornamerad(g+g^T)$?

0

$begingroup$

Is it true that, when we have a field $K$ with $operatornamechar(K) ne 2$, that $operatornamerad g = operatornamerad(g+g^T)$?

I know that in such a field, a bilinear form can be written as the sum of a symmetric and an alternating bilinear form.

How do I use this fact to prove the given statement?

Thanks.

linear-algebra

share|cite|improve this question

asked Mar 27 at 17:37

ZacharyZachary

3419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is rad?...
  $endgroup$
  – daw
  Mar 27 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @daw the radical
  $endgroup$
  – Zachary
  Mar 28 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $g$ is symmetric this is trivial since $g=g^T$, isn't it?
  $endgroup$
  – daw
  Mar 28 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So rad$g$=rad$(2g$)?
  $endgroup$
  – Zachary
  Mar 28 at 15:22
  

  
  
  
  

add a comment | 

0

$begingroup$

Is it true that, when we have a field $K$ with $operatornamechar(K) ne 2$, that $operatornamerad g = operatornamerad(g+g^T)$?

I know that in such a field, a bilinear form can be written as the sum of a symmetric and an alternating bilinear form.

How do I use this fact to prove the given statement?

Thanks.

linear-algebra

share|cite|improve this question

asked Mar 27 at 17:37

ZacharyZachary

3419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is rad?...
  $endgroup$
  – daw
  Mar 27 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @daw the radical
  $endgroup$
  – Zachary
  Mar 28 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $g$ is symmetric this is trivial since $g=g^T$, isn't it?
  $endgroup$
  – daw
  Mar 28 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So rad$g$=rad$(2g$)?
  $endgroup$
  – Zachary
  Mar 28 at 15:22
  

  
  
  
  

add a comment | 

$begingroup$

Is it true that, when we have a field $K$ with $operatornamechar(K) ne 2$, that $operatornamerad g = operatornamerad(g+g^T)$?

I know that in such a field, a bilinear form can be written as the sum of a symmetric and an alternating bilinear form.

How do I use this fact to prove the given statement?

Thanks.

linear-algebra

share|cite|improve this question

asked Mar 27 at 17:37

ZacharyZachary

3419

$endgroup$

share|cite|improve this question

asked Mar 27 at 17:37

ZacharyZachary

3419

share|cite|improve this question

asked Mar 27 at 17:37

ZacharyZachary

3419

asked Mar 27 at 17:37

ZacharyZachary

3419

asked Mar 27 at 17:37

ZacharyZachary

3419