Finding a base to a dual base This is a question about finding the base to a vector space which makes a specific dual space basis a dual basis.
I have a dual space $V^*$ (of a vector space $V$) with an ordered base $B^* = (y_1,\ldots,y_n)$.
I need to prove that there exist exactly one base $B=(x_1,\ldots,x_n)$ of $V$
which $B^*$ is the dual basis of $B$.
For the existence I know I need to find vectors $x_1,\ldots,x_n$ which $y_i(x_j)=\delta(i,j)$.
However I can't prove that there exist such vector, some hints?
 A: Pick any basis $(u_1, \dots, u_n)$ for $V$. For a general element $U=\sum_i a_i u_i$ of $V$, you have
$$y_j(U)=\sum_i a_i y_j(u_i).$$
Now suppose you want this to be $1$ when $j=1$, and $0$ otherwise. Then you get a system of $n$ linear equations in $n$ unknowns $a_1, \dots, a_n$, which will have a unique solution if and only if the associated matrix $[y_j(u_i)]_{ij}$ is invertible. I'll let you determine why this matrix is invertible. (Hint: what would the existence of a non-trivial zero linear combination of its rows imply? What happens if, for some $y \in V^*$, you have $y(u_i)=0$ for each $i$?)
A: Here's a different approach. Since this is homework, I only give an informal description with a few details missing; see if you can make this rigorous. 
Let $S = \{ y_1, \ldots, y_n \}$ and $S_{-i} = S \smallsetminus \{ y_i \}$. As a first step, 
we find $n$ dual vectors $z_1, z_2, \ldots, z_n$ such that 
$$
\langle z_i, y_j \rangle_{\ast} = \delta_{ij}.
$$ 
In fact, it is easy to write down what $z_i$ should be. Note that $\operatorname{Span}S_{-i}$ is an $(n-1)$-dimensional subspace orthogonal to $z_i$; so take the unit vector in the $1$-dimensional subspace $(\operatorname{Span} S_{-i} )^{\top}$ and normalize it suitably to make the inner product with $y_i$ equal to $1$. 
Once we get the vectors $z_i$, note that $z_i = x_i^{\ast}$ for some $x_i \in V$. 

Approach 2. In fact, this problem is trivial once you realise that $(V^*)^* = V$. Given  the basis $\{ y_1, y_2, \ldots, y_n \}$ for $V^*$, find the dual basis $B$ for $(V^\ast)^\ast = V$. Then you should be able to show that $B$ is the desired basis.
