Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Definition :Let $V$ be a finite dimensional real vector space. A para-complex structure on $V$ is an endomorphism $K$ : $V \to V$ such that:

  1. $K$ is an involution, that is $K^2 = Id_V$ ;
  2. The eigenspaces $V := ker(Id_V \mp K)$ of $K$ with eigenvalues $1$ respectively have the same dimension. A vector space $V$ endowed with a para-complex structure $K$, denoted by $(V;K)$, will be called para-complex vector space. We know, we can identify $K$ with $K=\pmatrix{Id_n & \cr & -Id_n\cr} $ . Is there a basis of vectors like {$ x_1,x_2,...,x_n,y_1,y_2,...,y_n $}, such that $Kx_i=y_i$ and $Ky_i=x_i$. i.e. can we write is such basis $K=\pmatrix{ & Id_n \cr Id_n & \cr} $ ?
share|improve this question

closed as off topic by Dmitri Pavlov, Qfwfq, Misha, Andreas Blass, Qiaochu Yuan May 12 '13 at 18:55

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

    
If you want the question deleted, you should ask The User for permission. –  S. Carnahan May 31 at 10:24

1 Answer 1

up vote 3 down vote accepted

$x_i=e_i+f_i, y_i=e_i-f_i$ where $e_1,\ldots,e_n,f_1,\ldots,f_n$ is the standard basis? Or what do you mean?

share|improve this answer
1  
I guess that this is too trivial—could you clarify? –  The User May 10 '13 at 22:59
2  
Btw, you are using $V$ for multiple different spaces I guess. –  The User May 10 '13 at 23:02
    
Yes, we just need this trick. because $Ke_i=e_i$ and $Kf_i=-f_i$ so we get the desired result.thanks "The User". –  Hassan Jolany May 10 '13 at 23:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.