# Show that $m\in M$ is unique

Let $$S$$ be a subspace of a normed linear space, $$X$$ and $$x_0\in X\backslash S$$. Consider the subspace spanned by $$M,$$ i.e. \begin{align} M:=[S\cup \{x_0\}]=\{m=x+\alpha\,x_0:\,x\in S,\;\text{for some}\;\alpha\in\Bbb{R}\} \end{align} I want to show that $$m$$ is unique.

MY TRIAL

Let $$m\in M,$$ then there exists $$\alpha\in\Bbb{R}$$ such that $$m=x+\alpha\,x_0.$$ Suppose we have another representation, then there exists $$\beta\in \Bbb{R},\;\beta\neq \alpha,$$ such that $$m=x+\beta\,x_0.$$ Thus, \begin{align} x_0=0 \;\text{which implies}\;x_0\in S,\;\text{contradiction, since }\; x_0\notin S.\end{align} Please, I'm I right? If not, could you please, provide an alternative proof?

• I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + \alpha x_0 = y + \beta x_0$, where $x, y \in S$, then $x = y$ and $\alpha = \beta$? – Theo Bendit Jan 10 at 5:01
• @Theo Bendit: That's true! That almost shows that I'm not correct! – Omojola Micheal Jan 10 at 5:02
• You're almost correct; instead consider $x - y = (\beta - \alpha)x_0$. – Theo Bendit Jan 10 at 5:03
• @Theo Bendit: Okay, let me try that! – Omojola Micheal Jan 10 at 5:04
• If you get, write an answer. :-) – Theo Bendit Jan 10 at 5:05

Let $$x_0\in X\backslash S$$ be arbitrary. Suppose that $$m\in M,$$ then there exists $$x\in S\;\text{and}\;\alpha\in\Bbb{R}$$ such that $$m=x+\alpha\,x_0.$$ Assume there is another representation, then there exists $$y\in S\;\text{and}\;\beta\in \Bbb{R}$$ such that $$m=y+\beta\,x_0,$$ where $$x\neq y$$ or $$\alpha\neq\beta$$. If $$x\neq y$$, then $$\beta \neq \alpha.$$ Otherwise, if $$\beta \neq \alpha$$ then $$x=y$$ or $$x\neq y.$$ In both cases, $$\beta \neq \alpha$$. Thus, \begin{align} \left(x+\alpha\,x_0=y+\beta\,x_0\right)&\implies (x-y)=(\beta-\alpha)\,x_0\\&\implies x_0=\dfrac{1}{\beta-\alpha}(x-y)\in S,\;\text{where}\;\beta-\alpha\neq 0\;\text{and}\;x-y\in S,\\&\implies x_0\in S,\;\text{since }\;S\;\text{is a subspace of a normed linear space }\\&\implies\text{a contradiction }\end{align} Hence, the representation $$m=x+\alpha x_0,\;$$ for $$\;m\in M$$ is unique.
• You should make the final implication more clear. What's wrong with $\alpha - \beta \neq 0$? You should be able to use this assumption to conclude $x_0 \in S$ (though you won't be able to conclude $x_0 = 0$). – Theo Bendit Jan 10 at 5:17