Functional in the dual space s.t $\psi$ restricted to a subspace equals $\phi$ Let $W$ be a subspace in a vector space $V$ [you may assume that dim $V$ is finite].
Prove that any $\phi$ in $W^*$ can be extended to a functional on $V$ , i.e. there exists $\psi$ in $V^*$ such that $\psi|W$ = $\phi$. ($\psi$ restricted to $W$ equals $\phi$). 
 A: Let $\,W_F=Span\{w_1,\ldots,w_k\}\,$ , and complete this basis of $\,W\,$ (i.e., a linear independent set) to a basis of $\,V_F\,:\;\;\{w_1,\ldots,w_k\,,v_1\ldots ,v_m\}\,$ . Define now
$$\Phi:V\to F\,\,,\,\,\Phi(w_i)=\phi(w_i)\;,\;\;\Phi(v_j)=k_j\,\,,\,k_j\in F\,\,,\,\forall\,\,i=1,...,k\,\,,\,\,j=1,...,m$$
and extend the definition by linearity to the whole of $\,V\,$ . Clearly $\,\left.\Phi\right|_W=\phi\,$ . Observe you have quite a few extensions for any such functional $\,\phi\,$ .
A: I don't understand, Doc, your problem with my "quite a few" comment in my answer. Shall we do an example? Take $\,W:=\left\langle\binom{1}{0}\right\rangle\,$ , and we complete the basis $\,\left\{\binom{1}{0}\right\}\,$ of $\,W\,$ to the basis $\,\left\{\binom{1}{0}\,,\,\binom{0}{1}\right\}\,$ of $\,\Bbb R^2\,$. 
Take for example $\,\phi\in W^*\,\,,\,\phi\binom{1}{0}:=1\,$ , and now I define 
$$\Phi_r\in(\Bbb R^2)^*\,\,,\Phi_r\binom{1}{0}:=1\,\,,\,\Phi_r\binom{0}{1}:=r$$
and extend by linearity:
$$\Phi_r\binom{x}{y}=\Phi_r\left(x\binom{1}{0}+y\binom{0}{1}\right)=x+ry$$
For example
$$\Phi_2\binom{1}{3}=1+3\cdot 2=7\;\;,\;\;\Phi_{-4}\binom{1}{3}=1+3(-4)=-11\,,\,etc.$$
How many extension you get? Of course, the codomain of all these non-zero functionals is the same, namely the whole of $\,\Bbb R\,$ , but as functions they all are different...Could it be you're confusing these two things? Of course, perhaps I'm missing something here, but I think the above explains what I meant.
