# rank of quadrics

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.)

• What exactly is the "rank of quadric"? The rank of the defining (symmetric) matrix? – Peter Franek Jun 28 '14 at 17:57
• @Peter: yes ${}$ – Georges Elencwajg Jun 28 '14 at 18:25

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 $$\Sigma_{n-r-1}$$.
• 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. – Georges Elencwajg Jun 28 '14 at 19:08