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Could you please help me to prove the following:

Let $f$ be any non-zero functional on a vector space $X$ and $x_0$ is a fixed element of $X\setminus N(f)$, where $N(f)$ is the kernel of $f$. Show that any $x\in X$ has a unique representation $x = ax_0 + y$, $y\in N(f)$.

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What have you tried? You have to prove two things: that every $x$ has such a representation, and that no $x$ has two such representations. Can you prove either? – Chris Eagle Jan 19 '13 at 16:31
Well I tried something but I couldnt understand where I failed.If you can,could you please send me the whole solution ? So that I can compare my solution. – Turku Kirli Jan 19 '13 at 16:39
  • If $x\in X$ then $x-\dfrac{f(x)}{f(x_0)}x_0\in N(f)$ (why?)...
  • If $x = a_1x_0 + y_1= a_2x_0 + y_2$ then $ (a_1-a_2)x_0 + (y_1-y_2)\in N(f)$ ...
    since $(ax_0\in N(f)\iff a=0) \Longrightarrow$ ...
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