Non degenerate implies even dimensional

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Given a finite-dimensional non-zero purely odd super-vector space with non-degenerate super-symmetric pairing $(-,-)$, that is $(x,y)=-(y,x)quadforall x,y$.



How does this imply that this super-vector space is even dimensional?




linear-algebra vector-spaces




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  • Did you try doing the one dimensional case? What did you get?
    – Somos
    Jul 22 at 13:52










  • If a basis is $x$, then $(x,x)=-(x,x)$, so $(x,x)=0$ and thus $(v,x)=0$ for all $v$, but $xneq 0$, contradiction to non-degenerate.
    – GIng
    Jul 22 at 13:58










  • Good work. Now try the two dimensional case. It will be interesting.
    – Somos
    Jul 22 at 14:00










  • Well there is nothing to show if the dimension is 2. Did you mean the three dimensional case?
    – GIng
    Jul 22 at 14:01










  • You are correct, technically. So try it next.
    – Somos
    Jul 22 at 14:03














up vote
1
down vote

favorite












Given a finite-dimensional non-zero purely odd super-vector space with non-degenerate super-symmetric pairing $(-,-)$, that is $(x,y)=-(y,x)quadforall x,y$.



How does this imply that this super-vector space is even dimensional?




linear-algebra vector-spaces




share|cite|improve this question





















  • Did you try doing the one dimensional case? What did you get?
    – Somos
    Jul 22 at 13:52










  • If a basis is $x$, then $(x,x)=-(x,x)$, so $(x,x)=0$ and thus $(v,x)=0$ for all $v$, but $xneq 0$, contradiction to non-degenerate.
    – GIng
    Jul 22 at 13:58










  • Good work. Now try the two dimensional case. It will be interesting.
    – Somos
    Jul 22 at 14:00










  • Well there is nothing to show if the dimension is 2. Did you mean the three dimensional case?
    – GIng
    Jul 22 at 14:01










  • You are correct, technically. So try it next.
    – Somos
    Jul 22 at 14:03












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Given a finite-dimensional non-zero purely odd super-vector space with non-degenerate super-symmetric pairing $(-,-)$, that is $(x,y)=-(y,x)quadforall x,y$.



How does this imply that this super-vector space is even dimensional?




linear-algebra vector-spaces




share|cite|improve this question













Given a finite-dimensional non-zero purely odd super-vector space with non-degenerate super-symmetric pairing $(-,-)$, that is $(x,y)=-(y,x)quadforall x,y$.



How does this imply that this super-vector space is even dimensional?





linear-algebra vector-spaces





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 22 at 13:35
























asked Jul 22 at 11:39









GIng

786




786











  • Did you try doing the one dimensional case? What did you get?
    – Somos
    Jul 22 at 13:52










  • If a basis is $x$, then $(x,x)=-(x,x)$, so $(x,x)=0$ and thus $(v,x)=0$ for all $v$, but $xneq 0$, contradiction to non-degenerate.
    – GIng
    Jul 22 at 13:58










  • Good work. Now try the two dimensional case. It will be interesting.
    – Somos
    Jul 22 at 14:00










  • Well there is nothing to show if the dimension is 2. Did you mean the three dimensional case?
    – GIng
    Jul 22 at 14:01










  • You are correct, technically. So try it next.
    – Somos
    Jul 22 at 14:03
















  • Did you try doing the one dimensional case? What did you get?
    – Somos
    Jul 22 at 13:52










  • If a basis is $x$, then $(x,x)=-(x,x)$, so $(x,x)=0$ and thus $(v,x)=0$ for all $v$, but $xneq 0$, contradiction to non-degenerate.
    – GIng
    Jul 22 at 13:58










  • Good work. Now try the two dimensional case. It will be interesting.
    – Somos
    Jul 22 at 14:00










  • Well there is nothing to show if the dimension is 2. Did you mean the three dimensional case?
    – GIng
    Jul 22 at 14:01










  • You are correct, technically. So try it next.
    – Somos
    Jul 22 at 14:03















Did you try doing the one dimensional case? What did you get?
– Somos
Jul 22 at 13:52




Did you try doing the one dimensional case? What did you get?
– Somos
Jul 22 at 13:52












If a basis is $x$, then $(x,x)=-(x,x)$, so $(x,x)=0$ and thus $(v,x)=0$ for all $v$, but $xneq 0$, contradiction to non-degenerate.
– GIng
Jul 22 at 13:58




If a basis is $x$, then $(x,x)=-(x,x)$, so $(x,x)=0$ and thus $(v,x)=0$ for all $v$, but $xneq 0$, contradiction to non-degenerate.
– GIng
Jul 22 at 13:58












Good work. Now try the two dimensional case. It will be interesting.
– Somos
Jul 22 at 14:00




Good work. Now try the two dimensional case. It will be interesting.
– Somos
Jul 22 at 14:00












Well there is nothing to show if the dimension is 2. Did you mean the three dimensional case?
– GIng
Jul 22 at 14:01




Well there is nothing to show if the dimension is 2. Did you mean the three dimensional case?
– GIng
Jul 22 at 14:01












You are correct, technically. So try it next.
– Somos
Jul 22 at 14:03




You are correct, technically. So try it next.
– Somos
Jul 22 at 14:03















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