Understanding first part of dual basis proof The textbook I'm reading attempts to proof the following: given $\left\{v_1, \ldots, v_n \right\}$ a basis for a vectorspace $V$ over $K$, there exists a basis $\left\{ \phi_1, \ldots, \phi_n \right\}$ for $V^{*}$ defined by \begin{align*} \phi_i (v_j) = \delta_{ij}. \end{align*} Here, each $\phi_i$ is a linear functional, i.e. an element of $V^{*}$. I understand the part of the proof where he shows linear independence, but I'm not sure if I can follow him where he proves these linear functionals span $V^{*}$. Here is what the author claims:

We first show that $\left\{ \phi_1, \ldots, \phi_n \right\}$ spans
  $V^{*}$. Let $\phi$ be an arbitrary element of $V^{*}$, and suppose
  $\phi(v_1) = k_1, \phi(v_2) = k_2, \ldots, \phi(v_n) = k_n$. Set
  $\sigma = k_1 \phi_1 + \ldots + k_n \phi_n$. Then \begin{align*}
\sigma(v_1) &= (k_1 \phi_1 + \ldots + k_n \phi_n) (v_1) \\ &= k_1
\phi_1 (v_1) + k_2 \phi_2 (v_1) + \ldots + k_n \phi_n (v_1) \\ &= k_1
\cdot 1 + k_2 \cdot 0 + \ldots + k_n \cdot 0 = k_1. \end{align*}
  Similarly for $i= 2, \ldots, n$. Since $\phi$ and $\sigma$ agree on
  the basisvectors, $\phi = \sigma = k_1 \phi_1 + \ldots k_n \phi_n$.
  Accordingly, $\left\{ \phi_1, \ldots, \phi_n \right\}$ spans $V^{*}$.

I got two questions/complaints. 1) Why does he set $\sigma = k_1 \phi_1 + \ldots + k_n \phi_n$ right away at the beginning? Is he allowed to do that? To me, this seems like circular reasoning. 
2) What does he mean with the assertion that $\phi$ and $\sigma$ agree on the basisvectors? Does he mean they determine the same action on the basis? I don't realy understand how he concludes from this that $\phi_1, \ldots, \phi_n$ span $V^{*}$. 
Some help/clarifications would be appreciated.
 A: An easier way to prove that the list $\phi_1, \dots, \phi_n$ spans $V^*$ is to note that $\dim V^* = n$ [more generally, the dimension of the space of linear maps from a finite-dimensional vector space $V$ to a finite-dimensional vector space $W$ equals $(\dim V)(\dim W)$]. You have already shown that $\phi_1, \dots, \phi_n$ is a linearly independent list in $V^*$. Because a linearly independent list of length $n$ in an $n$-dimensional vector space is a basis, we can conclude without further work that $\phi_1, \dots, \phi_n$ is a basis of $V^*$.
A: *

*The claim is that every linear functional $\varphi:V\to K$ is a (unique) linear combination of $\phi_1,\dots,\phi_n$. Then he sets a linear combination $\sigma$ of these so that it agrees with $\varphi$ on all the basis vectors $v_1,\dots,v_n$.

*Yes, this means that they determine the same action on the basis, i.e. $\varphi(v_i)=\sigma(v_i)$ for all $i$. And hence, as the $v_i$'s form a basis, by linearity of both $\varphi$ and $\sigma$, we will obtain that $\varphi(v)=\sigma(v)$ for all vectors $v\in V$, which finishes the proof (that $\varphi$ is a linear combination of $\phi_i$'s).

