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Consider the quadric $xw-yz$ in $\mathbf{P}^3$ (all over $\mathbf{C}$), and the Klein quadric $x_0 x_5+x_1 x_4+x_2 x_3$ in $\mathbf{P}^5$. I want to determine the rank of these quadrics. For the first I see that this quadric can be written as $s^T A s$ with $s=(x,y,z,w)$ and $A$ having entries $a_{14}=-a_{23}=-a_{32}=a_{41}=1/2$ and all other zero. From this I can conclude, since $\mathrm{rank} \ A=4$, that the first quadric has rank $4$ and similarily that the second one has rank $6$. So is this correct, both the result and the approach? Talking from experience, what would you say is the shortest way to determine the rank of a quadric? (I should note that I am mostly interested in a computational, i.e. linear algebra approach.)

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  • $\begingroup$ What exactly is the "rank of quadric"? The rank of the defining (symmetric) matrix? $\endgroup$ – Peter Franek Jun 28 '14 at 17:57
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    $\begingroup$ @Peter: yes ${}$ $\endgroup$ – Georges Elencwajg Jun 28 '14 at 18:25
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a) The rank of any smooth quadric in $\mathbb P^n(\mathbb C)$ is $n+1$.

b) If the quadric $Q$ is not smooth, an elementary algorithm due to Gauss (who else?) linearly transforms its equation to $x_0^2+\ldots +x_r^2=0\quad (r\lt n)$ and its rank is then $r+1$.
In that case the quadric $Q$ is a generalized cone:
$\bullet$ Its base is the smooth quadric $Q_0\subset \mathbb P^r(\mathbb C)=\{x_{r+1}=\ldots=x_n=0\}\subset \mathbb P^n(\mathbb C)$ given by the equations $x_{r+1}=\ldots=x_n=x_0^2+\ldots +x_r^2=0$
$\bullet \bullet$ Its generalized vertex is the linear subspace $ \Sigma_{n-r-1}=\{x_{0}=\ldots=x_r=0\}\subset \mathbb P^n(\mathbb C)$.
That linear subspace $ \Sigma_{n-r-1}\cong \mathbb P^{n-r-1}(\mathbb C) $ is exactly the set of singular points of the quadric $Q$ and it is the union of all the lines joining some point of $Q_0$ to some point of $ \Sigma_{n-r-1}$.

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  • $\begingroup$ First of all thank you for your answer. I agree with your first sentence, however I actually want to prove that my quadrics are smooth by computing the rank. Sorry, I should have said that, this is also why I emphasized that I am interested in a computational solution. How would you otherwise prove that these quadrics are smooth? $\endgroup$ – Zlatan der Zechpreller Jun 28 '14 at 19:00
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    $\begingroup$ Smoothness is very easy to check: just verify that the set of partial derivatives $\partial q/\partial x_i\; (i=0,...,n)$ of the homogeneous quadratic form $q$ defining the quadric have no non-trivial zero. This is elementary linear algebra since the partial derivatives are linear forms. $\endgroup$ – Georges Elencwajg Jun 28 '14 at 19:08
  • $\begingroup$ I see, I forgot about that this also works for projective varieties. But apart from that, is my own approach valid? It is also rather short $\endgroup$ – Zlatan der Zechpreller Jun 28 '14 at 19:14
  • $\begingroup$ Yes, your approach is valid. $\endgroup$ – Georges Elencwajg Jun 28 '14 at 19:36

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