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Let $Z$ be a proper subspace of an $n$ dimensional vector space $X$ and let $x_0\in X-Z$. Show that there exists a linear functional $f$ on $X$ such that $f(x_0)=1$ and $f(x)=0\forall x\in Z$.

Let $\{(z_i):1\leq i\leq m\};m<n$ be a basis of $Z$ .Then I can extend this linearly independent set to form a basis of $X$ say $\{(z_i):1\leq i\leq m\}\cup \{x_{m+1},x_{m+2}...x_n\}$.

Define $f:X\to K$ where $K$ is the scalar field of $X$ such that $f(z_i)=0;f(x_{m+1})=f(x_{m+2})=...=f(x_n)=1$.

Now if $x\in Z\implies x=\sum_{i=1}^m \beta_iz_i\implies f(x)=0$;

$x\in X-Z\implies x=\sum_{i=1}^m \beta_iz_i+\sum_{i={m+1}}^n\beta_ix_i\implies f(x)=\sum_{i={m+1}}^n\beta_if(x_i)$.

Now define $g:X\to K$ by $g(x)=\dfrac{f(x)}{|f(x)|}$ which is the required functional.Is this correct?

Jendrik Stelzner
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    It could be the case that the coefficients of $x_0$ in the basis sum to $0$. Best let $x_0$ be an element of the basis you construct. – Daniel Fischer Jan 09 '16 at 14:28

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Notice that $g(x) = f(x)/|f(x)|$ is no longer linear because $g(\lambda x) = \mathrm{sgn}(\lambda) g(x)$. Instead, notice that $\{z_i \mid 1 \leq i \leq m\} \cup \{x_0\}$ is linearly independent because $x_0 \notin Z$. So when extending $\{z_i \mid 1 \leq i \leq m\}$ to a basis of $X$ you can choose $x_{n+1} = x_0$ and then use $f$ itself.

Jendrik Stelzner
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