Finding dual basis for $\mathbb R^2$

Question

let $b_1 = (1,0)^T$ and let $b_2 = (1,1)^T$ and consider the set $B$={$b_1,b_2$}

I know B is a basis for $\mathbb R^2$, and for that matter an orthonormal basis. I need to find the dual basis of $B^*$ of B for ($\mathbb R^2$)^* and the dual basis of $B^*$ for ($\mathbb R^2$)^** but I don't even know what ($\mathbb R^2$)^** is? Any guidiance on these three problems would be greatly appreciated

Answer 1

Recall that a linear form $\varphi$ maps a vector $(x,y)$ to a number in the following way: $\varphi(x,y)=ax+by$. Hence $\varphi(x,y)=\langle (a,b),(x,y)$ or in short $\varphi(.)=\langle(a,b),.\rangle$.

As stated by Steel Stueber you are looking for two linear forms $e_i^*$ with $e_i^*(e_j)=\delta_{ij}$. Explicitly, if $e_1^*(.)=\langle (a,b),.\rangle$ we’re looking for a vector $(a,b)$ such that $$\langle(a,b),(1,0)\rangle=1\quad\text{and}\quad\langle(a,b),(1,1)\rangle=0.$$ A quick calculation reveals that $(a,b)=(1,-1)$, hence $e_1^*(x,y)=x-y$. Similar $e_2^*(x,y)=y$.

Answer 2

You can think about the dual of any vector space $V$ over a field $k$ as $\text{Hom}_k(V,k)$. So here you need to find a basis of $\text{Hom}_\mathbb{R}(\mathbb{R}^2,\mathbb{R})$. A linear map is determined by where it sends the basis vectors. You can check that if you have any vector space $V$ with basis $e_1,...,e_n$ then the dual basis $e_1^*,...,e_n^*$ are the maps such that $e_i^*(e_j)=\delta_{ij}$. For the double dual, try applying the same procedure to $(\mathbb{R}^2)^*$. Let me know if you need anything clarified

Answer 3

The basis $({\bf b}_1,{\bf b}_2)$ of $V={\mathbb R}^2$ is not orthonormal, but never mind. The dual basis $(\beta_1,\beta_2)$ of $V^*$ consists of two functionals $\beta_i:\>V\to{\mathbb R}$. These two functionals compute the coordinates of vectors ${\bf z}\in V$ with respect to the new basis $({\bf b}_1,{\bf b}_2)$.

When we write ${\bf z}=(x,y)$ with respect the standard basis then in this simple example we have $${\bf z}=x{\bf e}_1+y{\bf e}_2=(x-y)(1,0)+y(1,1)=(x-y){\bf b}_1+y {\bf b}_2\ .$$ It follows that $$\beta_1(x,y)=x-y,\qquad \beta_2(x,y)=y\ .$$