When is the linear map of plugging $m$ numbers in a polynomial is surjective? Let $n$, $m$ be positive integers and $V_n$ be the vector space of the polynomials of degree less than or equal to $n$ whose coefficients are complex numbers.  For $m$ complex numbers $a_1,\dots,a_m$, define a linear map $F:V_n\rightarrow\mathbb C$ as
$$
F(f) := (f(a_1),\dots,f(a_m)).
$$
I would like to find the necessary and sufficient condition for $F$ to be surjective.
Obviously, a necessary condition for it is that $m\le n$. I know that relative to a basis $(1,x,\dots,x^n)$ of $V_n$, $F$ can be presented as
a matrix
\begin{pmatrix}
1 & a_1 & \cdots & a_1^n\\
\vdots & \vdots & \ddots & \vdots\\
1 & a_m & \dots & a_m^n
\end{pmatrix}
However, I do not how to use this fact.
I would appreciate if you could help me solve this problem.
 A: The matrix you've typed is known as the Vandermonde matrix and it's known to have rank $m$ ($m\leq n+1$ as you've assumed) if the $a_i$'s are all distinct for $1\leq i\leq m$. If $m=n+1$, then it's nonsingular and its determinant is given by $\prod_{1\leq i<j\leq n+1} (a_j-a_i)$ (see the linked Wikipedia article for more details). 
Alternatively, as far as the original problem is concerned, complex polynomials of degree at most $n$ are completely determined by their values at $m$ complex numbers if $m\geq n+1$. (Can you prove this? Hint: a complex polynomial of degree $n$ has at most $n$ distinct complex roots.) In particular, the reasoning you've provided in your answer combined with this alternate proof shows that the Vandermonde determinant is non-zero. (Unfortunately, we can't get an explicit value by this approach but showing that it's non-zero itself is interesting and important!)
I hope this helps!
A: Hints: 
1) If certain inequalities are satisfied, and the numbers $a_i$ are distint, you should check that you can use polynomials like
$$
f_1=(z-a_2)(z-a_3)\cdots (z-a_m),\quad f_2=(z-a_1)(z-a_3)\cdots(z-a_m),\quad f_3=\cdots.
$$
2) If there are repetitions among the numbers $a_1,\ldots,a_m$, then you should be able to diagnose another problem.
