An example calculated in Principles of Algebraic Geometry of Griffiths and Harris'. On page 413 they write:

Example. We can now make a second computation for Chern classes of projective space. Let $X_0, \ldots , X_n$ be linear coordinates on $\mathbb{C}^{n+1}$, and let $\mathfrak{E}$ and $\pi_*$ be as in the Euler sequence above. Let $A=(\alpha_{ij})$ be an $(n+1)\times (n+1)$ matrix all of whose minors are distinct and nonzero, and consider the vector fields: $v_i = \mathfrak{E} (\alpha_{i0} X_0 , \ldots , \alpha_{in}X_n) = \pi_* \sum_j \alpha_{ij} X_i \frac{\partial}{\partial X_j}$.

They write that:

... we see that $v_1$ vanishes at $X\in \mathbb{P}^n$ exactly when $[\alpha_{10}X_0 , \ldots , \alpha_{1n}X_n] = [ X_0 , \ldots , X_n ] $

I do not understand how did they derive the last part, if we write the last identity then $v_1 = \pi_* \lambda \sum_j X_1 \frac{\partial}{\partial X_j}$ where $\lambda \in \mathbb{C}$, how do I see that $v_1$ vanishes from that identity?
Thanks in advance.
 A: $\newcommand{\dd}{\partial}\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$I don't have access to the book, but the indexing appears to be off. (There should be $X_{j}$ instead of $X_{i}$ on the right side of $v_{i}$.) Assuming that's correct:
If $A = [\alpha_{ij}]$ is an invertible $(n + 1) \times (n + 1)$ complex matrix, $X = (X_{0}, \dots, X_{n})$ denotes Cartesian coordinates in $\Cpx^{n+1}$ and $\dd = (\frac{\dd}{\dd X_{0}}, \dots, \frac{\dd}{\dd X_{n}})$ denotes the Cartesian frame in $T\Cpx^{n+1}$, then
$$
V_{i} := \sum_{j=0}^{n} \alpha_{ij}X_{j} \frac{\dd}{\dd X_{j}}
$$
defines a frame in $T\Cpx^{n+1}$ whose elements descend to $\Cpx\Proj^{n}$. Specifically, if $\pi:\Cpx^{n+1}\setminus\{0\} \to \Cpx\Proj^{n}$ denotes the projection, then the image $v_{i} := \pi_{*}(X)(V_{i})$ is independent of the representative $X$. For example, in the affine chart $\{X_{0} \neq 0\}$, we have $\pi(X) = \frac{1}{X_{0}}(X_{1}, \dots, X_{n})$, and a short calculation shows
$$
\pi_{*}(tX)(tV_{i}) = \pi_{*}(X)(V_{i})\quad\text{for all $t$ in $\Cpx^{\times}$.}
$$
Now, $\ker\pi_{*}(X)$ is spanned by the value of the "tautological vector field" $\sum X_{j} \frac{\dd}{\dd X_{j}}$ at $X$, precisely because $\pi$ sends each line through the origin to a point of $\Cpx\Proj^{n}$. Consequently, if $V_{i}$ is proportional to the tautological field along some line through the origin, then $v_{i} = 0$ at the corresponding point $[X]$ of $\Cpx\Proj^{n}$.
