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Let $V$ be a finite-dimensional vector space and $W\subset V$ a subspace. Show that each member of $W^{*}$ (the dual space) is a restriction of an element of $V^{*}$ in $W$

I tried to extend a linear functional defined on $W$ to a linear functional on $V$, but I failed because I couldn't get a LINEAR one. Is there another way that could be fertile?

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    $\begingroup$ Choose a complementary subspace $U$ to $W$ in $V$ and define the functional on $U$ to be zero. $\endgroup$ – amsmath Oct 11 '17 at 18:33
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    $\begingroup$ Use Hahn Banach theorem $\endgroup$ – Guy Fsone Oct 11 '17 at 18:35
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    $\begingroup$ do you assume finitedimensionality? $\endgroup$ – M. Van Oct 11 '17 at 18:37
  • $\begingroup$ @M.Van yes, I forgot it, sorry $\endgroup$ – Robson Oct 11 '17 at 18:41
  • $\begingroup$ @amsmath but this isn't linear, it is? I can't prove linearity if a take $w \in W$ and $u \in U$ (and verify if $f(w+u) = f(w) + f(u)$) $\endgroup$ – Robson Oct 11 '17 at 18:45
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Let $\varphi \in W^*$. Let $\{w_1,..,w_k\}$ be a basis of $W$. Extend this basis to a basis of $V$, $\{w_1,..,w_k,v_1,..,v_n \}$. Remember we can always uniquely define a linear map by giving the images of the basis elements (if you don't know this, check it!), so define $\psi : V \rightarrow \mathbb{R}$ by $\psi(w_i)=\varphi(w_i)$ and $\psi(v_i)=0$. Then $\varphi=\psi |_W$.

EDIT: So the statement I'd like you to check if you don't know it already is the following: Let $\{e_1,..,e_s \}$ be a basis for some vector space $U_1$, and let $a_1,..,a_s$ be elements in some other vector space $U_2$. Then there exists a unique linear map $T:U_1 \rightarrow U_2$ such that $T(e_i)=a_i$ for all $i$. In fancy terms, this is (a case of) the universal property for free modules.

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