Let us consider the short exact sequence of coherent sheaves on $\mathbb{P}^n$ $$0 \to \mathcal{O}\_{\mathbb P^n}(-1)^{r} \stackrel{N}{\longrightarrow} \mathcal{O}\_{\mathbb P^n}^{r} \longrightarrow \mathcal F \to 0,$$
where $r \geq 2$ and $N=\{L\_{ij}\}$ is a $r \times r$ matrix of linear forms in the variables $x\_0, \ldots, x\_n$ acting on the left (i.e., $\mathbf{x} \to N \mathbf{x}$). 


Then the sheaf $\mathcal{F}$ is supported on the determinantal hypersurface $Y:=\{\det N=0 \} \subset \mathbb{P}^n$, and it is locally free of rank $1$ (i.e., a line bundle) on the open dense locus of $Y$ given by $\textrm{rank } N = r-1$.


Passing to cohomology, we obtain an isomorphism $$\sigma \colon H^0(\mathbb{P}^n, \, \mathcal{O}\_{\mathbb P^n}^{r}) \stackrel{\cong}{\longrightarrow} H^0(Y, \, \mathcal F),$$
in particular $h^0(Y, \, \mathcal F)=r.$ In other words, this is a bijective correspondence between $n$-ples $(\alpha\_1, \ldots, \alpha\_r)$ of complex numbers and global sections of $\mathcal F$. 



> 
> **Question.** How can we describe the zero locus in $Y$ of the section 
>  $\sigma(\alpha\_1, \ldots, \alpha\_n)$ in terms of the $\alpha\_i$ and the matrix $N$?
> 
> 
> 




---


**Toy model.** Consider the case $n=3$, $r=2$ and $$N = \pmatrix{x\_0 & x\_1 \\ x\_2 & x\_3}.$$ 
In this situation, $Y$ is the smooth quadric $\{x\_0x\_3-x\_1x\_2=0 \} \subset \mathbb{P}^3$, and $\mathcal{F}$ is a line bundle with $2$ global sections, hence a pencil corrisponding to one of the two rulings of $Y$. I made some (unelegant) local computation and it seems to me that in this case $\sigma(\alpha\_1, \, \alpha\_2)$ should be something like $\alpha\_1 \sigma\_{02} + \alpha\_2 \sigma\_{13}$, where $\sigma\_{02}$ is the section whose zero locus is the line $\{x\_0=x\_2=0 \}$ and $\sigma\_{13}$ is the section whose zero locus is the line (in the same ruling) $\{x\_1=x\_3=0 \}.$ In other words, the global sections of $\mathcal{F}$ are expressed in terms of the *columns* of $N$.


On the other hand, dualizing our exact sequence and applying Grothendieck duality we get $$0 \to \mathcal{O}\_{\mathbb P^3}(-1)^{2} \stackrel{ { }^t N}{\longrightarrow} \mathcal{O}\_{\mathbb P^3}^{2} \longrightarrow \mathcal F^\*(1) \to 0,$$
where $\mathcal{F}^\*:=\mathscr{Hom}(\mathcal{F}, \, \mathcal{O}\_{\mathbb P^3})$.


Now the sections of $\mathcal{F}^\*(1)$ should correspond the other ruling of the quadric, because $\mathcal{F} \otimes \mathcal{F}^\*(1)= \mathcal{O}\_Y(1)$. Therefore the global sections of $\mathcal{F}^\*(1)$ are given in terms of the *rows* of $N$, that are the columns of ${}^t N$.



> 
> Is this correct? If so, is there any intrinsec or general way to see this? 
> 
> 
> 



Global sections of coherent sheaves on determinantal hypersurfaces in $\mathbb{P}^n$
This is meant to be an integration to Yusuf answer.


Consider a section 
$$
\stackrel{\rightarrow}{\alpha}=
\begin{pmatrix}
\alpha\_1\\
\vdots\\
\alpha\_r
\end{pmatrix} \in {\mathbb C}^r\cong H^0({\mathbb P}^n,{\mathcal O}^r\_{{\mathbb P}^n}).
$$
Its image on $H^0(Y,{\mathcal F})$ vanishes in a point $p \in Y$, by definition of quotient, if and only it is in the image of the map induced by $N$ on the stalks over $p$.
Then, by the Kronecker-Rouché-Capelli's Theorem, its zero locus equals the set of points where the rank of the matrix $N$ equals the rank of the $r \times (r+1)$ matrix $(N \stackrel{\rightarrow}{\alpha})$.


In your toy model, in all points of $Y$ the rank of $N$ is $1$, so we are looking to the locus where 
$$
rank
\begin{pmatrix}
x\_0&x\_1&\alpha\_1\\
x\_2&x\_3&\alpha\_2\\
\end{pmatrix}=1
$$
that is $\{\alpha\_1x\_2-\alpha\_2x\_0=\alpha\_1x\_3-\alpha\_2x\_1=0\}$.


This is as in your local computation indeed one of the rulings (but the other one, if my computations are not wrong).


A similar computation in the general case, as pointed out by Yusuf, produces equations involving the $(r-1) \times (r-1)$ minors of $N$.


More precisely, we need the adjoint or cofactor matrix $N^\*$ whose $(i,j)$ entry is $(-1)^{i+j}$ times the minor of $N\_{i,j}$, the determinant of the $(r-1) \times (r-1)$ matrix obtained by $N$ deleting the $i^{th}$ row and the $j^{th}$ column. In your toy model
$$
N^\*=
\begin{pmatrix}
x\_3&-x\_2\\
-x\_1&x\_0\\
\end{pmatrix}
$$


Then the zero locus of the image of $\stackrel{\rightarrow}{\alpha}$ is 
$$
^tN^\* \stackrel{\rightarrow}{\alpha}=0
$$


