Finding The Dual Basis Of Vector Space V2(R)

Jul 10, 2025 by ADMIN 45 views

$Decoding the Dual Basis of Vector Space V₂(\&R) with Basis B = { (2, 1), (3, 1) }$

Introduction: Understanding Vector Spaces and Bases

In the realm of linear algebra, vector spaces form the foundational structure upon which many mathematical concepts are built. A vector space, denoted as V, is essentially a collection of objects called vectors, which can be added together and multiplied by scalars. These operations must adhere to a specific set of axioms, ensuring the algebraic consistency of the space. The quintessential example of a vector space is the n-dimensional Euclidean space, often represented as &Rⁿ, where vectors are n-tuples of real numbers. In our specific case, we are dealing with V₂(&R), which represents the two-dimensional Euclidean space, where vectors are ordered pairs of real numbers, such as (a, b).

A basis of a vector space is a set of linearly independent vectors that span the entire space. This means that any vector in the space can be expressed as a unique linear combination of the basis vectors. Linear independence ensures that no vector in the basis can be written as a linear combination of the others, while spanning guarantees that the basis vectors can "reach" every point in the vector space. The basis provides a coordinate system for the vector space, allowing us to represent vectors as unique combinations of the basis elements. For V₂(&R), a common basis is the standard basis {(1, 0), (0, 1)}, but as the problem states, B = {(2, 1), (3, 1)} also forms a valid basis. This means any vector in V₂(&R) can be written as a unique linear combination of (2, 1) and (3, 1). Exploring different bases provides valuable insights into the structure of vector spaces and their transformations.

Delving into Dual Spaces and Dual Bases

The concept of a dual space is pivotal in understanding the deeper structure of vector spaces. Given a vector space V over a field F (in our case, the field of real numbers &R), the dual space, denoted as V*, is the set of all linear transformations from V to F. These linear transformations are often called linear functionals. A linear functional f : V → F is a function that satisfies the properties of linearity: f(u + v) = f(u) + f(v) and f(cv) = cf(v) for all vectors u, v in V and scalar c in F. In simpler terms, a linear functional takes a vector as input and produces a scalar output, while preserving the linear structure of the vector space.

The dual basis is a basis for the dual space V* that is intimately related to the original basis of V. If B = {v₁, v₂, ..., vₙ} is a basis for V, then the dual basis B* = {f₁, f₂, ..., fₙ} is a set of linear functionals in V* that satisfy a specific orthogonality condition: fᵢ(vⱼ) = 1 if i = j, and fᵢ(vⱼ) = 0 if i ≠ j. This condition essentially states that each functional in the dual basis "picks out" the corresponding vector in the original basis, returning 1, and annihilates all other basis vectors by returning 0. The dual basis provides a powerful tool for analyzing linear transformations and understanding the relationship between a vector space and its dual.

Finding the Dual Basis for B = { (2, 1), (3, 1) }

Now, let's tackle the core problem: finding the dual basis for B = {(2, 1), (3, 1)} in V₂(&R). We are looking for two linear functionals, f₁ and f₂, that satisfy the dual basis condition. This means we need to find f₁ and f₂ such that:

f₁(2, 1) = 1
f₁(3, 1) = 0
f₂(2, 1) = 0
f₂(3, 1) = 1

Since f₁ and f₂ are linear functionals on V₂(&R), we can express them in the general form: f₁(a, b) = c₁a + d₁b and f₂(a, b) = c₂a + d₂b, where c₁, d₁, c₂, and d₂ are real numbers. Our goal is to determine these coefficients such that the orthogonality conditions are met.

Let's apply the conditions to f₁:

f₁(2, 1) = 2c₁ + d₁ = 1
f₁(3, 1) = 3c₁ + d₁ = 0

Subtracting equation (1) from equation (2), we get: c₁ = -1. Substituting this value back into equation (1), we have: 2(-1) + d₁ = 1, which gives us d₁ = 3. Therefore, f₁(a, b) = -a + 3b.

Now, let's apply the conditions to f₂:

f₂(2, 1) = 2c₂ + d₂ = 0
f₂(3, 1) = 3c₂ + d₂ = 1

Subtracting equation (1) from equation (2), we get: c₂ = 1. Substituting this value back into equation (1), we have: 2(1) + d₂ = 0, which gives us d₂ = -2. Therefore, f₂(a, b) = a - 2b.

Thus, the dual basis of B = {(2, 1), (3, 1)} is {f₁, f₂}, where f₁(a, b) = -a + 3b and f₂(a, b) = a - 2b.

Verification and Significance of the Dual Basis

To ensure the correctness of our result, we can verify that the obtained linear functionals indeed satisfy the dual basis conditions:

f₁(2, 1) = -(2) + 3(1) = 1
f₁(3, 1) = -(3) + 3(1) = 0
f₂(2, 1) = (2) - 2(1) = 0
f₂(3, 1) = (3) - 2(1) = 1

These calculations confirm that the functionals f₁ and f₂ form the dual basis of B. The dual basis plays a crucial role in various areas of mathematics and physics. For instance, it is used in tensor analysis, differential geometry, and quantum mechanics. The dual basis provides a way to represent linear functionals as vectors, allowing us to perform algebraic manipulations on them. It also provides a connection between a vector space and its dual, which is essential for understanding concepts like orthogonality and adjoint operators.

Conclusion: Mastering Dual Bases in Linear Algebra

In conclusion, we have successfully determined the dual basis of the vector space V₂(&R) with respect to the basis B = {(2, 1), (3, 1)}. The dual basis, {f₁, f₂}, consists of the linear functionals f₁(a, b) = -a + 3b and f₂(a, b) = a - 2b. This process involved understanding the definitions of vector spaces, bases, dual spaces, and dual bases. By applying the orthogonality condition, we were able to solve for the coefficients of the linear functionals and obtain the desired dual basis. The concept of dual bases is a fundamental tool in linear algebra, providing a powerful way to analyze linear transformations and understand the structure of vector spaces. Mastering dual bases opens doors to advanced topics in mathematics and physics, making it an essential concept for anyone pursuing studies in these fields. Understanding these concepts is not just about solving problems; it's about developing a deeper appreciation for the elegant structures that underpin the mathematical world. By grasping the intricacies of vector spaces and their duals, we gain a more profound understanding of the language in which the universe is written. Thus, the journey into the dual space is not merely an exercise in computation, but a step towards a more complete and nuanced understanding of the mathematical fabric that surrounds us.