Proof of $\text{Span}(A \cup \{x_0\})$ for $x_0 \notin A$ can uniquely determine $x = m + ax_0$? I have a difficult time understanding the claim $x = m + ax_0$ as stated in below text (it regards the proof of the Hahn-Banach theorem). I would rather look at a proof or a beginning of a proof; could anyone help me out a bit? It would be really appreciated.

 A: If $x\in \operatorname{Span}(M \cup \{x_0\})$, then $$x = \sum_{i = 1}^k \lambda_i m_i + \alpha x_0$$ for some scalars $\lambda_1,\ldots, \lambda_k, \alpha$ and vectors $m_1,\ldots, m_k\in M$. If $m = \sum \lambda_i m_i$, then $m\in M$ since $M$ is a subspace of $E$. Thus $x = m + \alpha x_0$ with $m\in M$.
Suppose $x = m + \alpha x_0 = m' + \alpha' x_0$ for some $m,m'\in M$ and scalars $\alpha, \alpha'$. Then $(\alpha - \alpha')x_0 = m' - m \in M$; the condition $x_0\in E\setminus M$ forces $\alpha = \alpha'$. Then $m = m'$. Hence, the representation in the proof is unique.
A: The condition $x_0 \in E \setminus M$ basically means that $M$ and $\operatorname{span}\{x_0\}$ are linearly independent, and then the condition $E = \operatorname{span}(M \cup \{x_0\})$ means that $E$ is the internal direct sum of $M$ and $\operatorname{span}\{x_0\}$, so each element of $E$ can be represented uniquely as an element of $M$ plus an element of $\operatorname{span}\{x_0\}$, i.e. as $m + ax_0$ where $m \in M$ and $a \in \Bbb R$.
A: To elaborate this post a little bit; I feel like that there is a need for a proof of $x_0$ being linearly independent to $M$, so here is my suggestion;

Corollary. for any $m \in M$ and $x_0 \in X-M$, the vectors $x_0$ and $m$ are linearly independent.

Proof. the elements (vectors) $m$ and $x_0$ are linearly independent if $am + bx_0=0$ implies $a,b=0$. Suppose $b \neq 0$ and $am + bx_0=0$. It will be shown that this leads to a contradiction.
As $b \neq 0$, the equality $am + bx_0 = 0$ is equivalent to $-\frac{am}{b}=x_0$. Furthermore, over the defined field $a,b \in \mathbb{F}$ we have $-a \in \mathbb{F}$ and $-\frac{a}{b} \in \mathbb{F}$. By the "linear" properties of the space $M$, it yields that $m'=-\frac{am}{b} \in M$.
This leads to a contradiction, as $m'=x_0$ but $m' \in M$ and $x_0 \notin M$.
