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Let $v\in \mathbb{C}^n$ be an $n$ dimensional complex vector. Define the non-standard bilinear form $\left< u,v \right> = u^T v$ (the usual inner product except without the conjugation). What are the properties of the self orthogonal vectors, those such that $\left < v,v \right>=0$? What are the properties of this space of vectors?

For example, if we say that $v^Tv = 0$ has the null property, then so does $\lambda v$ for any complex $\lambda$, and so does $v^*$ (the conjugated vector). Clearly, the set of null vectors is not a vector space since generally $u+v$ is not null if $u$ and $v$ are null.

This may be well known, but I just have no idea what to Google for. Any pointers to existing literature are appreciated.

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    $\begingroup$ These vectors are called isotropic vectors (for the complex symmetric bilinear form you mentioned). $\endgroup$ – Marc van Leeuwen Nov 30 '11 at 7:13
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The most natural way to view your set of vectors is in the setting of projective geometry; the vector space $K^n$ (where $K$ is now an arbitrary field, so $K = \mathbb{C}$ for you) can be seen as a projective space $\mathbb{P}^{n-1}(K)$ of dimension $n-1$. (The "points" of the projective space $\mathbb{P}^{n-1}(K)$ correspond to the vector lines $\{ \lambda v \mid \lambda \in K \}$ of $K^n$).

In your example, the set of isotropic vectors for the bilinear form $\langle u, v \rangle = u^T v$ correspond to a quadric in $\mathbb{P}^{n-1}(K)$. See, for instance, http://en.wikipedia.org/wiki/Quadric_(projective_geometry). Note that over an algebraically closed field such as $\mathbb{C}$, there is essentially only one non-degenerate bilinear form, so also only one quadric.

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