Finding Dual Basis from a basis in $\mathbb R^2$ I am given the fact that these two vectors form the basis B of $\mathbb R^2$:
$$
B=\{\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix}\}
$$
and then asked to find the dual basis or the basis of $(\mathbb R^2)^*$. 
I would really appreciate if anyone could explain the procedure to follow for these types of questions.
 A: So, I'm hoping you know the definition of dual bases: given a basis $(v_1, v_2)$ for $\mathbb{R}^2$, it's a basis $(f_1, f_2)$ for $(\mathbb{R}^2)^*$ such that $f_1(v_1) = f_2(v_2) = 1$ and $f_1(v_2) = f_2(v_1) = 0$.
Recall that $(\mathbb{R}^2)^*$ is essentially $\mathbb{R}^2$. Every linear functional $f$ on $\mathbb{R}^2$ can be expressed by:
$$f(x, y) = x f(1, 0) + y f(0, 1) = (x, y) \cdot (f(1, 0), f(0, 1)),$$
which allows us to linearly and bijectively map $f \in (\mathbb{R}^2)^*$ to $(f(1, 0), f(0, 1)) \in \mathbb{R}^2$. Using this vector to find the image of $(x, y)$ under $f$ is as simple as taking the dot product. So, we can turn our attention to finding vectors $w_1, w_2 \in \mathbb{R}^2$ such that
$$v_1 \cdot w_1 = v_2 \cdot w_2 = 1 \text{ and } v_1 \cdot w_2 = v_2 \cdot w_1 = 0.$$
Now, there's a simple way to construct such vectors. If you can find a matrix $A$ such that
$$A \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$$
then the rows of $A$ are the dual basis you need. Why? Just think about matrix multiplication. To get the top left entry of the product (i.e. $1$), you take the dot product of the first row of $A$ (i.e. $w_1$) with the first column of $\begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix}$, i.e. $v_1$. Similar observations show that the rows of $A$ must be the dual basis.
That is, the dual basis will be the rows of
$$\begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix},$$
which is to say, $((-1, 3), (1, -2))$. As linear maps,
\begin{align*}
f_1(x, y) &= -x + 3y \\
f_2(x, y) &= x - 2y.
\end{align*}
You should check to make sure it satisfies the dual basis definition.
This method extends out of $\mathbb{R}^2$ of course. It will just result in a larger matrix for you to invert.
A: $(\mathbb{R}^2)^*$ consists of linear maps $\ell:\mathbb{R}^2\to \mathbb{R}$ which  have  standard matrix representation 
$$ \begin{bmatrix}
a&b
\end{bmatrix}$$
where $a=\ell(e_1)$ and $b=\ell(e_2)$ for $e_1,e_2$ the  standard basis. Let's write your basis vectors as $b_1,b_2$ respectively. The dual basis is the basis $(\phi_1,\phi_2)$ for $(\mathbb{R}^2)^*$ satisfying $\phi_i(b_j)=\delta_{ij}$ for $\delta_{ij}=1$ with $i=j$ and $0$ else. So, $\phi_1(b_1)=1$ and $\phi_1(b_2)=0$. Notice that $b_2-b_1=e_1$. So, 
$$\phi_1(e_1)=\phi_1(b_2-b_1)=\phi_1(b_2)-\phi_1(b_1)=-1.$$
Similarly, $3b_1-2b_2=e_2$ so
$$ \phi_1(e_2)=\phi_1(3b_1-2b_2)=3\phi_1(b_1)-2\phi_1(b_2)=3.$$
Hence, $\phi_1$ has matrix representation
$$ \begin{bmatrix}
-1&3
\end{bmatrix}.$$
Applying the same reasoning, we get 
$$ \phi_2(e_1)=\phi_2(b_2-b_1)=\phi_2(b_2)-\phi_2(b_1)=1$$
$$ \phi_2(e_2)=\phi_2(3b_1-2b_2)=3\phi_2(b_1)-2\phi_2(b_2)=-2.$$
Hence, $\phi_2$ has matrix representation 
$$ \begin{bmatrix}
1&-2
\end{bmatrix}.$$
A simple calculation reveals that indeed these $\phi_1,\phi_2$ satisfy $\phi_i(b_j)=\delta_{ij}$.
A: You want a pair of vectors $v^1,v^2$ such that $v^iv_j=\delta^i_j=\begin{cases}1, i=j\\0, i\neq j\end{cases}$.
So, $v^1=(-1,3),v^2=(1,-2)$ would work.
For $v^1$ you get the system $\begin{cases}  2x+y=1\\3x+y=0\end{cases}$, and similarly for $v^2$.
