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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
asked Jul 22 at 11:39
GIng
786
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1
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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
asked Jul 22 at 11:39
GIng
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
 |Â
show 2 more comments
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
asked Jul 22 at 11:39
GIng
786
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
asked Jul 22 at 11:39
GIng
786
edited Jul 22 at 13:35
asked Jul 22 at 11:39
GIng
786
asked Jul 22 at 11:39
GIng
786
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
 |Â
show 2 more comments
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
 |Â
show 2 more comments
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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