Extending linear function from a subspace to the whole (finite-dimensional) space Let $V$ be a finite-dimensional vector space over the field $F$ and let $W$ be a subspace of $V$. If $f$ is a linear functional on $W$, prove that there is a linear functional g on $V$ such that $g(\alpha)$ = $f(\alpha)$ for each $\alpha$ in the subspace $W$.
 A: Choose a basis $w_1, w_2, \ldots, w_k$ for $W$, where $k = \dim W$.  Extend this to a basis of $V$ by adding linearly independent vectors $v_{k + 1}, v_{k + 2}, \ldots, v_n$, where $n = \dim V$ and $v_j \notin W$ for $k < j \le n$.  Set $g(w_j) = f(w_j)$ for $1 \le j \le k$ and set $g(v_j) = 0$ for $k < j \le n$.  This defines $g$ on the basis $\{w_1, w_2, \ldots, w_k, v_{k + 1}, v_{k + 2}, \ldots, v_n \}$.  Extend $g$ to all of $V$ by linearity:  if $x \in V$ is given (uniquely!) by $x = \sum \alpha_j w_j + \sum \beta_k v_k$ then we set $g(x) = \sum \alpha_j g(w_j) + \sum \beta_k g(v_k)$.  It is easy to see that $g$ so defined is a linear functional on $V$.  Furthermore since in fact $g(x) = \sum \alpha_j g(w_j) = \sum \alpha_j f(w_j$) for $x \in W$, we see that $g$ and $f$ agree on W.  QED.
Nota Bene: It is worth noting that the $g(v_k)$ may in fact be taken arbitrarily in the above construction; choosing $g(v_k) = 0$ is merely a convenience.  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Let $\;\{u_1,...,u_k\}\; $ be a basis of $\;W\;$, and complete it to a basis $\;\{u_1,...,u_k,u_{k+1},..,u_n\}\;$ of V. Now define
$$g(u_i):=\begin{cases}f(u_i)&,\;\;1\le i\le k\\{}\\0&,\;\;k+1\le i\le n\end{cases}$$
and extend the definition by linearity.
A: The linear functional on the subspace is defined by its values on the basis of the subspace. This basis can be constructed in a way that makes it subset of the basis of the whole space. Then how can you extend your functional to the whole space? Just define some values for your functional on the rest of the basis. 
It is apparent that this choice of values on the whole basis is not unique. Thus the extension is not unique.
