How to prove the linear map is surjective? I am stuck with the proof of Goldstine theorem, here is the step I cannot understand. Let $X$ be a vector space, and $f_i:X\to \mathbb{C}$ ($1\leq i\leq n$) be linear functionals which is linearly independent, then the linear map $X\to\mathbb{C}^n$ defined by
$$ x\mapsto (f_1(x),\dots,f_n(x))$$ is surjective
is surjective. How to show this? 
 A: Let $f(x) = (f_1(x),...,f_n(x))$ and $S=\{ f(x) | x \in X \} \subset \mathbb{C}^n$.
If $S$ is a proper subspace of $\mathbb{C}^n$, there is some non zero $\alpha \in \mathbb{C}^n$ such that $ \alpha \bot S$. 
Then $\alpha^* f(x) = 0$ for all $x$, or $\sum_k \overline{\alpha_k} f_k (x) = 0$ for all $x$ or $\sum_k \overline{\alpha_k} f_k = 0$, which implies that the
$f_k$ are linearly dependent.
Hence $S$ cannot be a proper subspace of $\mathbb{C}^n$ and hence 
$f$ is surjective.
Alternative: This approach avoids dealing with orthogonal complements
but is more tedious and (I think) is less intuitive.
Let $N = \cap_k \ker f_k$ and $Y = X /N$. Note that if $x_1-x_2 \in N$, then $f(x_1-x_2) = 0$ and so $f(x_1) = f(x_2)$ and so 
$\tilde{f}([x]) = f(x)$ is well defined. Note that $\ker \tilde{f}$ is trivial and ${\cal R} f = {\cal R} \tilde{f}$.
Then $\dim X/N \le n$. To see this, suppose there are $n+1$ linearly independent $y_i = [x_i] \in Y$, and let
$A= \begin{bmatrix} \tilde{f}(y_1) & \cdots & \tilde{f}(y_{n+1}) \end{bmatrix}$, since $A: \mathbb{C}^{n+1} \to \mathbb{C}^n$ we see that
there must be some $\beta \neq 0$ such that $A \beta = 0$ , or
$\sum_k \beta_k \tilde{f}(y_k) = \tilde{f}(\sum_k \beta_k y_k) = 0$,
and so $\sum_k \beta_k y_k = 0$ (that is, $N$), hence a contradiction.
Let $y_1,...,y_m$ be a basis for $X/N$ and define
$F=\begin{bmatrix} \tilde{f}(y_1) & \cdots & \tilde{f}(y_{m}) \end{bmatrix}$. Then note that ${\cal R} F = {\cal R} \tilde{f}$. It
is clear that ${\cal R} F \subset {\cal R} \tilde{f}$, and since
$y_1,...,y_m$ is a basis, any $y$ can be written $y=\sum_k \beta_k y_k$ and hence $f(y) = F \beta$ so we have equality.
Now note that the vectors $(\tilde{f}_k(y_1),...,\tilde{f}_k(y_m))$ (the rows of $F$)
are linearly independent. To see this, suppose
$\sum_k \beta_k (\tilde{f}_k(y_1),...,\tilde{f}_k(y_m)) = 0$. Then
$\sum_k \beta_k \tilde{f}_k(y_j) =0$ for all $j$ and hence
$\sum_k \beta_k \tilde{f} = 0$ and so $\beta_k = 0$. Hence $F$ has
at least (in fact, exactly) $n$ linearly independent columns and
hence ${\cal R } F = \mathbb{C}^n$.
