Equivalent characterizations of rational normal curve A rational normal curve $C \subset \mathbb{P}_k^d$ (assume $k= \mathbb{C}$) can be defined usually up to projective equivalence in two equivalent ways:

*

*smooth irreducible nondegenerate curve $C \subset \mathbb{P}^d$ of minimal degree $\text{deg}(C)=d$


*projectively equivalent to the image of the Veronese map
$$ \nu_d: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \
[X_0:X_1] \mapsto [X_0^d: X_0^{d-1}X_1:..., X_1^d]  $$
The second definition can be reprased as that there exist a $\mathbb{C}$-basis of the vector space of homogeneous polynomials $p_0,p_1,..., p_d$ of degree $d$ in two variables $X_0, X_1$ such that $C$ is the image under
$$ P: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \
[X_0:X_1] \mapsto [p_0(X): p_1(X):..., p_d(X)]  $$
In Harris' book Algebraic Geometry, on page 14 there is an alternative presumably equivalent construction of a rational normal curve:
Start by choosing $d$ codimension two linear
spaces $\Lambda_i \cong \mathbb{P}^{d-2} \subset \mathbb{P}^{d}$. The family $\{H_i(\lambda)\}$ of hyperplanes in $\mathbb{P}^{d}$ containing $\Lambda_i$ is
then parameterized by $\lambda \in \mathbb{P}^1$; choose such parameterizations, subject to the condition
that for each $\lambda$ the planes $H_1(\lambda), ... , H_d(\lambda)$ are independent, i.e., intersect in a
point $ p(\lambda)$. Claim: It is then the case that the locus of these points $p(\lambda)$ as $\lambda$ varies in $\mathbb{P}^1$ is a rational normal curve.
Question: Why this construction gives a rational normal curve $\bigcup_{\lambda \in \mathbb{P}^{1}} H_1(\lambda) \cap ... \cap H_d(\lambda)$?
I found here a rather promising approach discussing identical problem: in the linked discussion is showed how one can construct explicitly the point $p(\lambda)= [p_0(\lambda): p_1(\lambda):...: p_d(\lambda)]$ in terms of certain homogenous polynomials $p_i(\lambda)$ of degree $d$ in coordinates $(\lambda)=\lambda_0, \lambda_1$.
The idea is quite simple. Consider the $(d+1) \times (d+1)$ matrix $A(\lambda)$ where each $j$-th row of $A(\lambda)$ is encoded by the $j$-th family of hyperplanes as follows. The parametrized family $H_j(\lambda)$ of hyperplanes which contain $\Lambda_i \cong \mathbb{P}^{d-2}$ is given as vanishing loci of $\lambda_0S_j +\lambda_1T_j$, where
the two linear forms $S_j, T_j$ in  variables $Y_0,..., Y_d$ determine $\Lambda_j$. The $j$-th row of $A(\lambda)$ is encoded in
$$a_0(\lambda)Y_0+ a_1(\lambda)Y_1 +...+ a_d(\lambda)Y_d=  \lambda_0S_j +\lambda_1T_j   $$
The $(d+1)$-row is filled with zeroes. The adjugate matrix $P(\lambda)$ having $d$-minors of $A$ as entries satisfies $A \cdot P=P \cdot A= \text{det}(A) \cdot E_{d+1} $ and $\text{det}(A) $ is zero. The $(d+1)$-th column of $P(\lambda)$ is the unique point $p(\lambda)=[p_0(\lambda): p_1(\lambda):...: p_d(\lambda)]$ with homogeneous $p_i(\lambda)$ of degree $d$.
One expects finally to obtain the rational normal curve as image  of
$$[X_0:X_1] \mapsto [p_0(X): p_1(X):..., p_d(X)]  $$
but it isn't clear why the $p_i(\lambda)$ are linearly independent. If they aren't, we obtain a degenerate curve in the image, that's not what we want.
 A: Let $\mathbb{P}^d = \mathbb{P}(V)$. A hyperplane in $\mathbb{P}(V)$ corresponds to an epimorphism $V \to k$, a pencil of hyperplanes to an epimorphism
$$
V \otimes \mathcal{O}_{\mathbb{P}^1} \to \mathcal{O}_{\mathbb{P}^1}(1),
$$
and a collection of $d$ pencils to a morphism
$$
f \colon V \otimes \mathcal{O}_{\mathbb{P}^1} \to \mathcal{O}_{\mathbb{P}^1}(1)^{\oplus d}.
$$
The condition imposed on the pencils means that this is an epimorphism, therefore its kernel $\mathrm{Ker}(f)$ is a line subbundle in $V \otimes \mathcal{O}_{\mathbb{P}^1}$, and computing its determinant it is easy to see that
$$
\mathrm{Ker}(f) \cong \mathcal{O}_{\mathbb{P}^1}(-d).
$$
The embedding $\mathrm{Ker}(f) \to V \otimes \mathcal{O}_{\mathbb{P}^1}$ induces a degree $d$ map $\mathbb{P}^1 \to \mathbb{P}(V)$, and its image is precisely the curve in question. By this construction, it is a rational normal curve of degree $d$.
EDIT. To show that the curve is nondegenerate, consider the exact sequence
$$
0 \to 
\mathcal{O}_{\mathbb{P}^1}(-d) \to 
V \otimes \mathcal{O}_{\mathbb{P}^1} \stackrel{f}\to
\mathcal{O}_{\mathbb{P}^1}(1)^{\oplus d} \to 
0
$$
obtained above. Dualizing it, we obtain
$$
0 \to 
\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus d} 
\stackrel{f^\vee}\to
V^\vee \otimes \mathcal{O}_{\mathbb{P}^1} \to
\mathcal{O}_{\mathbb{P}^1}(d) \to 
0.
$$
Since the sheaf $\mathcal{O}_{\mathbb{P}^1}(-1)$ has no cohomology, the induced morphism
$$
V^\vee = 
H^0(\mathbb{P}^1, 
V^\vee \otimes \mathcal{O}_{\mathbb{P}^1}) \to
H^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(d))
$$
is an isomorphism. This means that the curve is nondegenerate.
