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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.

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$(\mathbb{R}^2)^*$ 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 $\phi_i(b_j)=\delta_{ij}$.

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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.

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  • $\begingroup$ @mr_e_man Thanks. I knew I was going to make an error like that as I was typing it out. $\endgroup$ – Theo Bendit Jan 24 at 1:44
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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 $v^2$.

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