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.


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.
