Answer: To compute the dual basis E' = (f¹, f²) of the given basis E = [(1,2), (0,1)] of R², we need to find two linear functionals f¹ and f² that satisfy the following conditions:
1. The dual basis vectors f¹ and f² should be linear functionals, which means they should map vectors from R² to the scalar field R.
2. The dual basis vectors should satisfy the condition that f¹(e₁) = 1 and f¹(e₂) = 0, and f²(e₁) = 0 and f²(e₂) = 1.
Let's calculate the dual basis vectors step by step:
First, we find f¹:
For f¹(e₁) = 1, we have f¹(1,2) = 1. Since f¹ is a linear functional, we can write f¹(x,y) = a₁x + b₁y, where a₁ and b₁ are scalars. Plugging in the values (1,2) for (x,y), we get the equation:
a₁(1) + b₁(2) = 1
Simplifying this equation, we have:
a₁ + 2b₁ = 1
Similarly, for f¹(e₂) = 0, we have f¹(0,1) = 0. Plugging in the values (0,1) for (x,y), we get the equation:
a₁(0) + b₁(1) = 0
Simplifying this equation, we have:
b₁ = 0
Solving the equations a₁ + 2b₁ = 1 and b₁ = 0, we find that a₁ = 1/2 and b₁ = 0. Therefore, the first component of the dual basis vector f¹ is f¹(x,y) = (1/2)x.
Next, we find f²:
For f²(e₁) = 0, we have f²(1,2) = 0. Similar to f¹, we can write f²(x,y) = a₂x + b₂y. Plugging in the values (1,2) for (x,y), we get the equation:
a₂(1) + b₂(2) = 0
Simplifying this equation, we have:
a₂ + 2b₂ = 0
Similarly, for f²(e₂) = 1, we have f²(0,1) = 1. Plugging in the values (0,1) for (x,y), we get the equation:
a₂(0) + b₂(1) = 1
Simplifying this equation, we have:
b₂ = 1
Solving the equations a₂ + 2b₂ = 0 and b₂ = 1, we find that a₂ = -2 and b₂ = 1. Therefore, the second component of the dual basis vector f² is f²(x,y) = -2x + y.
To summarize, the dual basis E' = (f¹, f²) of the basis E = [(1,2), (0,1)] of R² is given by:
f¹(x,y) = (1/2)x
f²(x,y) = -2x + y