# 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.

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

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.

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

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 $$v^2$$.