Hahn-Banach and normed space I have the next exercise: 
Let $E$ be a normed space, $n\in \mathbb{N}$ and $\{x_1,\dots,x_n\}\in E$ a family of independent vectors. Show that for all $\alpha_1,\dots,\alpha_n\in \mathbb{K}$ there is a linear form $T\in E'$ such that $T(x_i)=\alpha_i$, for all $i=1,\dots,n$.
My solution:
I used the following theorem derived from Hahn Banach's Theorem
Theorem (Bounded linear functionals): Let $X$ be a normed space and let $x_0\neq 0$ be any element of $X$. Then there exists a bounded linear functional $T$ on $X$ such that
$||T||=1$, $T(x_0)=||x_0||$.
How $\{x_1,\dots,x_n\}\in E$ a family of independent vectors, then each $x_i\neq 0$, with $i=1,\dots,n$. In addition $X$ is a normed space.
Therefore exists $T\in E'$ such that $||T||=1$, $T(x_i)=||x_i||$. Let $\alpha_i=||x_i||$. With which you have to $T(x_i)=\alpha_i$
 A: Use the following version of Hahn-Banach Extension Theorem (An Introduction to Banach Space Theory, Robert Megginson, page 75):

Suppose that $f_{0}$ is a bounded linear functional on a subspace $Y$ of a normed space $X$. Then there is a bounded linear functional $f$ on all of $X$ such that $\|f\|=\|f_{0}\|$ and the restriction of $f$ to $Y$ is $f_{0}$.

Now $Y=\left<x_{1},...,x_{n}\right>$ is finite-dimensional subspace of $X$, any linear functional on $Y$ is bounded. Given $\alpha_{i}$, we let $f_{0}(x_{i})=\alpha_{i}$ and extends canonically on the whole $Y$.
Applying Hahn-Banach Extension Theorem for an $f\in X^{\ast}$ such that $f\big|_{Y}=f_{0}$, then we are done.
A: The result you are trying to prove requires $\alpha_1,\ldots,\alpha_n$ be arbitrary, but at the end of your proof you have $\alpha_i=\|x_i\|$.  Thus your solution is incorrect.
In fact, the result your using (the corollary of the Hahn-Banach theorem), while it is a great one, isn't strong enough to show what you're trying to show.  What you should instead use is the following corollary of the Hahn-Banach theorem:

Let $X$ be a normed space, and let $M\subset X$ be a closed proper subspace.  If $x_0\in X\setminus M$, there is some $f\in X^*$ such that $f(x)=0$ for all $x\in M$ and $f(x_0)\neq 0$.  

