I posted in MO a few days ago a question about the equivalence of three characterizations of a rational normal curve $C \subset \mathbb{P}_k^d$ (assume $k= \mathbb{C}$):
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
$$ \nu_d: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \ [X_0:X_1] \mapsto [X_0^d: X_0^{d-1}X_1:..., X_1^d] $$
Projectively equivalence in the second characterization can be equivalently expressed as that there exist linearly independent $d+1$ homogeneous polynomials $p_0,p_1,..., p_d$ of degree $d$ in variables $X_0, X_1$ such that $C$ is the image of
$$ P: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \ [X_0:X_1] \mapsto [p_0(X): p_1(X):..., p_d(X)] $$
(this means just that that $p_0,p_1,..., p_d$ form $\mathbb{C}$-basis.) But there is also another geometric characterization of rational normal curve I found in Harris' book Algebraic Geometry, on page 14:
Choosing $d$ codimension two linear
spaces $\Lambda_i \cong \mathbb{P}^{d-2} \subset \mathbb{P}^{d}$. The pencil $\{H_i(\lambda)\}$ 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)$. Then
- $\bigcup_{\lambda \in \mathbb{P}^{1}} H_1(\lambda) \cap ... \cap H_d(\lambda)$ is a rational normal curve
In the linked MO discussion Sasha showed with cohomological methods that the last construction gives a rational normal curve.
I would like to know if possible to justify with an "elementary" geometric argument that the curve in third characterization is nondegenerate, i.e. is not contained in a hyperplane.
If we follow in approach from linked discussion then it's rather easy to show that the curve $\bigcup_{\lambda \in \mathbb{P}^{1}} H_1(\lambda) \cap ... \cap H_d(\lambda)$ arises as image of map
$$ [\lambda_0:\lambda_1] \mapsto [p_0(\lambda): p_1(\lambda):..., p_d(\lambda)] $$
with $p(X)$ homogeneous of degree $d$ in $\lambda_0, \lambda_1$. (Sketch: Let $\Lambda_j \cong \mathbb{P}^{d-2}$ be defined as vanishing set of two linear forms $S_j, T_j$ in variables $Y_0,..., Y_d$. Then the pencil $H_j(\lambda)$ of hyperplanes is given by vanishing loci of $\lambda_0 \cdot S_j +\lambda_1 \cdot T_j$. Form the $d \times (d+1)$ matrix $A(\lambda)$ whose $j$-th row is encoded in equation
$$a_0(\lambda)Y_0+ a_1(\lambda)Y_1 +...+ a_d(\lambda)Y_d= \lambda_0S_j +\lambda_1T_j $$
It is easy to see that for every $\lambda$ the unique (up to scalar) solution is given by $p(\lambda)=[\hat{A}_0(\lambda): \hat{A}_1(\lambda):...: \hat{A}_d(\lambda)]$ where $\hat{A}_i(\lambda) $ is the determinant of matrix obtained from $A(\lambda)$ removing $i$-th column. In short: the entries of the solution vector $p(\lambda)$ are $d$-minors of $A(\lambda)$. By construction every $\hat{A}_i $ is homogeneous of degree $d$ in $\lambda_0, \lambda_1$ and give rise to a map
$$ P: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \ [\lambda_0:\lambda_1] \mapsto [\hat{A}_0(\lambda): \hat{A}_1(\lambda):... \hat{A}_d(\lambda)] $$
That's our map.) The problem is that at this stage is not clear if the curve in the image is nondegenerate.
As remarked this can be shown with cohomological methods. and (multi) linear algebra methods (see the sketh below), but I hoped that it is also possible to show that this curve is nondegenerate with pure geometric considerations showing directly it's not possible that every all points $p(\lambda) = H_1(\lambda) \cap ... \cap H_d(\lambda)$ are contained in a hyperplane.
Here a linear algebra argument: We use the Fact that if $V$ is a vector space and $v_1, v_2,.., v_s$ are linear independent vectors in $V$, then for $d \le s$ the vectors $v_{i_1} \wedge v_{i_2} \wedge ... \wedge v_{i_d}$, $ \ i_1 < i_2 < ... <i_d$ are linear independent in $\bigwedge^dV$ too.
Therefore in order to show that the $d$-minors $\hat{A}_0(\lambda), \hat{A}_1(\lambda),..., \hat{A}_d(\lambda)$ of $A(\lambda)$ are linearly independent, we have to show that the column vectors of $A(\lambda)$ which we regard as vectors of vector space $W \otimes V$ where $W= \mathbb{C} \cdot \lambda_0+\mathbb{C} \cdot \lambda_1$ and $V=\mathbb{C}^d$.
If these column vectors are linearly dependent, then application of appropriate operations on columns of $A(\lambda)$ which can be realized by a multiplication from the right by a $GL_{d+1}(\mathbb{C})$-matrix would give us a new matrix $A'(\lambda)$ with $(d+1)$-th column vector equals zero.
Geometrically this transformation is nothing but a change of coordinates $Y_0,..., Y_d$, so it not changes the class of resulting map $\mathbb{P}^1 \to \mathbb{P}(V)$ under projective equivalence and $A'(\lambda)$ retains maximal rank $d$. But then for every $\lambda$ the solution equals $p(\lambda)=[0:...:1]$, a case which we excluded. Therefore the columns of $A(\lambda)$ are linearly independent and we win by the Fact.
Final question: Is it possible to give a pure geometric argument that the union $\bigcup_{\lambda \in \mathbb{P}^{1}} H_1(\lambda) \cap ... \cap H_d(\lambda)$ cannot be contained in a hyperplane. Even if previous arguments prove it, I still can't grasp intuitively the geometrical picture why such curve cannot be contained in a hyperplane and missing a bit a better "geometric insight" into the structure of this curve. Is it possible to give a pure geometric argument that this curve is nondegenerate? Is it also possible to deduce that this curve has degree $d$ directly from construction of the curve by counting intersections argument with general hyperplane instead of dipping in concrete algebraic equations?
One way to see non-degeneracy is the following. Consider the Segre embedding. $$ \mathbb{P}^d \times \mathbb{P}^1 = \mathbb{P}(V) \times \mathbb{P}(W) \subset \mathbb{P}(V \otimes W) = \mathbb{P}^{2d + 1}. $$ Then each pencils corresponds to a hyperplane (which I will also denote by $H_j$) in the ambient space. Let $$ C = (\mathbb{P}(V) \times \mathbb{P}(W)) \cap H_1 \cap \dots \cap H_d. $$ The assumption about the pencils is equivalent to the assumption that $p_W \colon C \to \mathbb{P}(W)$ is an isomorphism, so that $C \cong \mathbb{P}^1$, and the corresponding curve is just $$ p_V(C) \subset \mathbb{P}(V). $$ Thus, the question is why $p_V(C)$ is not contained in a hyperplane $\mathbb{P}(V') \subset \mathbb{P}(V)$, or why $C$ is not contained in $\mathbb{P}(V') \times \mathbb{P}(W)$.
Assume it is. Note that the Segre variety is nondegenerate, and a hyperplane section of a nondegenerate variety is nondegenerate. Therefore, $C$ is a nondegenerate curve, and its linear span has codimension $d$. In particular, any hyperplane containing $C$ is a linear combination of $H_j$.
Now the assumption we made means that any hyperplane containing $\mathbb{P}(V') \times \mathbb{P}(W)$ is a linear combination of $H_j$. Note that any such hyperplane intersects the Segre variety along a reducible set $$ (\mathbb{P}(V') \times \mathbb{P}(W)) \cup (\mathbb{P}(V) \times \mathbb{P}(W')),\tag{*} $$ where $W' \subset W$ is a hyperplane. Therefore, $C$ is an intersection of $(*)$ with $d-1$ hyperplanes, in particular, it contains an intersection of $$ \mathbb{P}(V) \times \mathbb{P}(W') = \mathbb{P}^d $$ with $d-1$ hyperplanes. Thus, $C$ contains a vertical line, which contradicts to the fact that $p_W$ is an isomorphism on $C$.