Dimension of the space of skew-symmetric bilinear functions

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SUMMARY

The dimension of the space of skew-symmetric bilinear functions on a vector space \( V \) of dimension \( n \) is definitively \( \frac{n(n-1)}{2} \). This conclusion arises from the properties of skew-symmetry, where \( f(u,v) = -f(v,u) \). The discussion highlights the importance of choosing a basis \( (e_1, \dots, e_n) \) for \( V \) and utilizing linearity to analyze the parameters involved in defining these functions.

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  • Understanding of linear algebra concepts, particularly bilinear forms.
  • Familiarity with vector spaces and their dimensions.
  • Knowledge of skew-symmetric functions and their properties.
  • Ability to work with basis representations in vector spaces.
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  • Study the derivation of the dimension formula for bilinear forms in linear algebra.
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smile1
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Hello everyone,

I stuck on this problem:
find the dimension of the space of dimension of the space of skew-symmetric bilinear functions on $V$ if $dimV=n$.

I thought in this way, for skew-symmetric bilinear functions, $f(u,v)=-f(v,u)$, then the dimension will be $n/2$

Am I right?

thanks
 
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Re: dimension of the space of skew-symmetric bilinear functions

smile said:
I thought in this way, for skew-symmetric bilinear functions, $f(u,v)=-f(v,u)$, then the dimension will be $n/2$
Do you think the space of skew-symmetric functions is a fractal when $s$ is odd? :)
Suppose $(e_1,\dots,e_n)$ is a basis of $V$, $u=\sum_{i=1}^nx_ie_i$ and $v=\sum_{i=1}^ny_ie_i$. Expand $f(u,v)$ using linearity and skew symmetry. If $x_i$ and $y_i$ are fixed, how many parameters do you need to set to get a concrete number for $f(u,v)$?
 

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