Finding a Dual Basis $B^*$ Let $S$ be the standard basis of $\Bbb R^2$ and let $S^*=\{x_1, x_2\}$ be it's dual basis.  Let $B=\{(1, 2),(4,7)\}$ be another basis of $\Bbb R^2$.  Compute $B^*$ in terms of $x_1, x_2$
I don't particularly understand.  I know for example that for $f_i(b_j)=1$ if $i=j$ and $0$ otherwise.  But farther than that I don't really get it.  
I think I have to find $2$ vectors for $B^*$, that follow $f_1(b^*_1)=1$, and so on following the above form.  I get lost going beyond this idea of what a dual basis is though
Any help, tips or resources to go look at is appreciated.  Thanks
 A: Following the definition that you've given, you should be able to verify that the functions $x_i$ on vectors $(v_1,v_2)$ are defined by
$$
x_1(v_1,v_2) = v_1\\
x_2(v_1,v_2) = v_2
$$
Similarly, if $y_i$ are the elements of $B^*$, we have
$$
y_1(v_1,v_2) = a_1 v_1 + b_1 v_2\\
y_2(v_1,v_2) = a_2 v_1 + b_2 v_2
$$
where $a_i,b_i$ are chosen so that the $y_i$ satisfy the definition of the dual basis.  
For $y_1$, we have
$$
y_1(1,2) = a_1 + 2b_1 = 1\\
y_1(4,7) = 4a_1 + 7b_1 = 0
$$
Solve this system to find that $a_1 = -7, b_1 = 4$.  In particular, this means that
$$
y_1(v_1,v_2) = -7x_1(v_1,v_2) + 4x_2(v_1,v_2)
$$
That is, $y_1 = -7x_1 + 4x_2$ is an expression for $y_1$ in terms of $x_1,x_2$.  Do the same for $y_2$, and you will have completely answered the question.
A: First calculates the dual basis of $B$ and then put it as a linear combination of $(x_{1},x_{2})$
A: You need to find $f_i$ such that $f_i(b_j) = \delta_{ij}$.
I use $e_1^*, e_2^*$ to denote the standard dual basis.
You will have $f_1 = a e_1^* + be_2^*$, $f_2 = c e_1^* + de_2^*$.
Then $f_i(b_j) = \delta_{ij}$ will give four equations in $a,b,c,d$. Solve
them.
Note that this is the same as solving
$\pmatrix{a & b \\ c & d} \pmatrix{1 & 4 \\ 2 & 7} = \pmatrix{1 & 0 \\ 0 & 1 }$.
Solution:

 You will get $f_1 = -7 e_1^* + 4 e_2^*$, $f_2 = 2 e_1^* -e_2^*$.

