Exercise 1.24 from Joe Harris' Algebraic Geometry: First Course I have a question about solvability of Exercise 1.24 (p. 14) from Joe Harris' Algebraic Geometry: A First Course and correctness of following 'synthetic' construction which according to the book (or alternatively due to page 79 in this script) should give us a rational normal curve. (Since we want to work as geometric as possible, all constructions and spaces are defined over $\mathbb{C}$)
Harris wrote:

As indicated, we can generalize this to a construction of rational normal curves
in any projective space $\mathbb{P}^d$. Specifically, 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 $p(\lambda)$. 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.


Exercise 1.24. Verify the last statement

So our constructed curve $C$ is given by
$$ C:= \bigcup_{\lambda \in \mathbb{P}^{1}} H_1(\lambda) \cap ... \cap H_d(\lambda)  $$
On page 10 Harris gave the standard definition of a rational normal
curve: A rational normal curve is defined as the image
of the map $v_d: \mathbb{P}^1 \to \mathbb{P}^d; 
\lambda \mapsto [A_0(\lambda):...: A_d(\lambda)]$
with an arbitrary basis $A_0, ... , A_d$ of the space of
homogeneous polynomials of degree $d$ on $\mathbb{P}^1 $.
Note that this curve is projectively equivalent to the image of the Veronese map $[X_0:X_1] \mapsto [X_0^d: X_0^{d-1}X_1:..., X_1^d]$.
The exercise:
Suppose we start by choosing $d$ codimension two linear
spaces $\Lambda_i \cong \mathbb{P}^{d-2} \subset \mathbb{P}^d, 
i=1,...,d$ (here I doubt that the $\Lambda_i$ can be choosen really arbitrary!) and consider $d$ families of hyperplanes
$\{H_i(\lambda) \}_{\lambda \in \mathbb{P}^1}, i=1, ..., d$
where each $ H_i(\lambda)$ contains $\Lambda_i$
parametrized by the line $\mathbb{P}^1$ such that for each
$\lambda \in \mathbb{P}^1$ the planes
$H_1(\lambda),..., H_d(\lambda)$ are linear independent,
i.e. intersect in a point $p(\lambda)$.
That is we have to check that the constructed curve $C$ above can be realized as
the image of such map $v_d$, that means that the map
$$ p: \mathbb{P}^1 \to \mathbb{P}^d,
\lambda \to p(\lambda)=  H_1(\lambda) \cap ... \cap H_d(\lambda) $$
has the form $\lambda \to [A_0(\lambda):...: A_d(\lambda)]$ for
appropriate basis $A_0,..., A_d$ of
the space of homogeneous polynomials of degree $d$ on $\mathbb{P}^1$.
The way I tried to solve it contains a serious problem, see below. We parametrize every linear space $\Lambda_i$ as vanishing set of two independent
linear polynomials $L_i, M_i \in \mathbb{C} \cdot Z_0 + \mathbb{C} \cdot Z_1... + 
\mathbb{C} \cdot Z_d $. So $\Lambda_i= V(L_i, M_i)$. Then every our pencil
$\{H_i(\lambda) \}_{\lambda \in \mathbb{P}^1}$ can be parametrized as vanishing
set
$$ \lambda_0 L_i + \lambda_1 M_i =0 $$
with running $\lambda=[\lambda_0: \lambda_1]$. If we reordner the terms after
variables $Z_i$ we obtain $d$ equitions
$$   \lambda_0 L_i + \lambda_1 M_i = 
r_{i,Z_0} (\lambda)Z_0+ r_{i,Z_1} (\lambda)Z_1 + ...
+ r_{i, Z_d} (\lambda)Z_d =0  $$
We can naturaly encode the coefficents of $i$-th family as $i$-th row
of following matrix $R \in Mat_{d+1}(\mathbb{C}[\lambda])$; the $(n+1)$-row
we fill with zeroes:
$$ \begin{pmatrix}
r_{1,Z_0} (\lambda) & r_{1,Z_1} (\lambda) & ... & r_{1,Z_d} (\lambda) \\
r_{2,Z_0} (\lambda) & r_{2,Z_1} (\lambda) & ... & r_{2,Z_d} (\lambda)\\
\\ ...  \\
\\ ...  \\
r_{d,Z_0} (\lambda) & r_{d,Z_1} (\lambda) & ... & r_{d,Z_d} (\lambda)\\
0 & 0 & ... & 0 \\
\end{pmatrix} $$
Since for every $\lambda$ the $H_1(\lambda), ..., H_d(\lambda)$ are independent,
this matrix $A$ has rank $d$.
What we do next looks promissing at first glance; we consider the
adjugate matrix $A$ of $R$, that's a unique matrix
$A \in Mat_{d+1}(\mathbb{C}[\lambda])$ with
$$ AR=RA= det(R) \cdot I_{n+1} = 0   $$
By definition of adjugate matrix it's $d+1$-th column $A_{\cdot, n+1}$ provides
exactly the solution we are looking for: for every $\lambda$ it is up to
multiplication by a scalar $c \neq 0$ the solution of linear equations
encoded by matrix $R$ and every entry of this column is by construction
a homogeneous polynomial of degree $d$ in $\lambda_0, \lambda_1$, let
set
$$  (A_0(\lambda), ... , A_d(\lambda)):= A_{\cdot, n+1}^T $$
and pass to it's homogenization.
Problem (a really serious one): Why are the $A_i $ linearly independent,
equivalently why they form a basis of the space of homogeneous deg $d$ polynomials
in $\lambda_0, \lambda_1$? (see also the comment by lhl73 in this closely related question dealing with nondegeneracy of curves of this type separately)
Indeed, that's equivalent to the property that the image of
$\lambda \mapsto [A_0(\lambda):...: A_d(\lambda)]$ is not contained in
a proper hyperplane $H \cong \mathbb{P}^{d-1} \subset \mathbb{P}^{n-1}$.
And if we looking again at the construction, Harris nowhere imposed
additional assumptions how the $\Lambda_i$ are related to each other;
it doesn't rule out eg the bad case $\Lambda_1= \Lambda_2$, since
one can find still two families $\{H_i(\lambda)\}, i=1,2$ of hyperplanes
containing $\Lambda_1 (=\Lambda_2)$ with $H_1(\lambda) \neq H_2(\lambda)$ for
every $\lambda$, so the imposed condition is not violated. But eg if we choose $\Lambda_1= \Lambda_2$ by construction the complete curve $C$ would be
comtained in $\Lambda_1$, therefore the $A_i(\lambda)$ constructed as above
will cannot be linearly independent and the curve will not be rational normal
curve as defined by Harris on page 10.
Question: Does anybody have experience with this Exercise 1.24 and
knows how to solve it correctly, or if it's indeed true that the
quoted construction not always gives a rational normal curve in sense of
Harris book? (there is also nowhere Errata of this book available)
Probably Harris also has forgotten to impose an additional assumption
on the spaces $\Lambda_i$ in the construction above or I'm just too stupid to solve the exercise & understand the construction. If that's so, can the construction be slightly modified in a most general way when one would obtain always a rational normal curve? E.g. does the construction give us always rational normal curve if we additionally require that all $\Lambda_i$ should be distinct?
In any case I would be very thankful if anybody who has experience with this
Exercise and construction would share how it can be solved correctly.
 A: I think here is a full solution why the polynomials
$A_0(\lambda),..., A_d(\lambda)$ must be
linearly independent and therefore build basis of
$n$-forms on $\mathbb{P}^1$ (and that solves
the "serious problem")
Recall that for every fix $\lambda=[\lambda_0: \lambda_1]$
the point $p_{[\lambda_0: \lambda_1]} \in \mathbb{P}^d$ is the
unique solution of independent linear system
$$ \lambda_0 L_1 + \lambda_1 M_1 = ... = \lambda_0 L_d + \lambda_1 M_d= 0 $$
We reorder the terms after variables $Z_0, Z_1, ..., Z_d$ and obtain $d$
equations
$$ \lambda_0 L_i + \lambda_1 M_i = 
r_{i,Z_0} (\lambda)Z_0+ r_{i,Z_1} (\lambda)Z_1 + ...
+ r_{i, Z_d} (\lambda)Z_d =0 $$
We naturaly encode the coefficents of $i$-th family as $i$-th row
of following matrix $R \in Mat_{d+1}(\mathbb{C}[\lambda])$; the $(n+1)$-row
we fill with zeroes:
$$ \begin{pmatrix}
r_{1,Z_0} (\lambda) & r_{1,Z_1} (\lambda) & ... & r_{1,Z_d} (\lambda) \\
r_{2,Z_0} (\lambda) & r_{2,Z_1} (\lambda) & ... & r_{2,Z_d} (\lambda)\\
\\ ...  \\
\\ ...  \\
r_{d,Z_0} (\lambda) & r_{d,Z_1} (\lambda) & ... & r_{d,Z_d} (\lambda)\\
0 & 0 & ... & 0 \\
\end{pmatrix} $$
There exist the adjugate matrix $A$ of $R$, that's a unique matrix
$A \in Mat_{d+1}(\mathbb{C}[\lambda])$ with
$$ AR=RA= det(R) \cdot I_{n+1} = 0   $$
By definition of adjugate matrix it's $d+1$-th column $A_{\cdot, n+1}$ provides
exactly the solution we are looking for: for every $\lambda$ it is up to
multiplication by a scalar $c \neq 0$ the solution of linear equations
encoded by matrix $R$ and every entry of this column is by construction
a homogeneous polynomial of degree $d$ in $\lambda_0, \lambda_1$, let
set
$$  (A_0(\lambda), ... , A_d(\lambda)):= A_{\cdot, n+1}^T $$
and pass to it's homogenization.
That's what we already know. Now the interesting part is why are $A_d(\lambda)$
$\mathbb{C}$-linearly independent. Assume the $A_0(\lambda), ... , A_d(\lambda)$
are lineraly dependent. Then there exist a $G \in GL_{n+1}(\mathbb{C})$ with
$$(G \cdot A_{\cdot, n+1})^T=(\widetilde{A_0}, ...,\widetilde{A_{d-1}}, 0)$$
where $G$ is the conposition of finitely many elementary row manipulation of $A$.
Because $\operatorname{adj}(AB)= \operatorname{adj}(B)\operatorname{adj}(A)$ we conclude
that $\widetilde{A}:= \frac{1}{det(G)}GA$ is the
adjugate matrix of $\widetilde{R}:=RG^{-1}$.
Since $\widetilde{R}$ differs from $R$ by finitely many column manipulations, we have
$\widetilde{R}(\lambda)=$
$$ \begin{pmatrix}
\widetilde{r}_{1,1} (\lambda) & \widetilde{r}_{1,2}(\lambda) & ... & \widetilde{r}_{1,d+1}(\lambda) \\
\\ ...  \\
\\ ...  \\
\widetilde{r}_{d,1} (\lambda)  & \widetilde{r}_{d,2} (\lambda)  & ... & \widetilde{r}_{1,d+1} (\lambda) \\
0 & 0 & ... & 0 \\
\end{pmatrix} $$
and by definition of adjugate matrix we obtain $\operatorname{det}(\widetilde{R}_{d+1, d+1})=0$
for the $(d+1, d+1)$-minor of $\widetilde{R}$.
Therefore there exist a $Z:=[z_0: z_1:...: z_{d-1}:0] \in \mathbb{P}^d$ with $\widetilde{R}(\lambda) \cdot Z=0$
for all $\lambda \in \mathbb{P}^1$ simultaneously. Then our map $p$ is constant, a contradiction.
Therefore the $A_i$ are linearly independent.
Last remark: How changes $p$ due to the left action of $G^{-1}$ of $R$. Well, this action changes
the columns of $R$ and therefore changes just the coordinates $Z_0, Z_1,..., Z_d$; that means
it maps the image under $p$ to a projectively equivalent curve. Since every projectively equivalent
curve of a rational normal curve is rnc as well by definition, this action of column of
not changes the content of the statement and we are done.
