I'm asked what is the smallest possible dimension of a vector space with a non-degenerative skew-symmetric bi-linear form.

I realized it could not be size one because then $\delta_{ij}=-\delta_{ij}$, since, taking the skew-symmetric bi-linear form $\tau$, for it to be skew-symmetric $\langle\tau\small(\vec {v}\small),\vec v\rangle=-\langle(\vec {v},\tau \small(\vec v\small)\rangle$.

Then it can be size two with the condition $\tau\small(\vec v)=-\vec u$, and $\tau\small(\vec u)=-\vec v$ where $S=\{ \vec v,\vec u\}$ is a basis of my vector space.

Anyway I´m not sure the proof is right.


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