Basis And Dimension For Subspaces W₁ And W₂ In ℝ⁵
Introduction
In linear algebra, understanding vector spaces and their subspaces is crucial. A subspace is a subset of a vector space that is itself a vector space, satisfying the same axioms. To fully characterize a subspace, we often seek a basis, which is a set of linearly independent vectors that span the subspace, and the dimension, which is the number of vectors in a basis. This article delves into finding a basis and dimension for two specific subspaces, W₁ and W₂, of the vector space ℝ⁵. These subspaces are defined by linear equations that constrain the components of the vectors within them. We will explore the methodologies to identify a basis for each subspace and determine their respective dimensions, providing a comprehensive understanding of their structure within ℝ⁵.
The process involves expressing the given conditions as linear equations and then solving them to express some variables in terms of others. This allows us to parameterize the vectors in the subspace, revealing the linearly independent vectors that form the basis. The number of these independent vectors then gives us the dimension of the subspace. This approach is fundamental in linear algebra and is used extensively in various applications, including solving systems of linear equations, understanding linear transformations, and in fields like computer graphics, data analysis, and physics. By the end of this exploration, you will have a solid understanding of how to find the basis and dimension of subspaces defined by linear constraints, empowering you to tackle similar problems in linear algebra with confidence.
(a) Subspace W₁ = {(a₁, a₂, a₃, a₄, a₅) ∈ ℝ⁵ | a₁ - a₃ - a₄ = 0}
Understanding Subspace W₁
To find a basis and dimension for the subspace W₁, we need to analyze the given condition: a₁ - a₃ - a₄ = 0. This equation defines a hyperplane in ℝ⁵, which is a subspace. To find a basis, we need to express the vectors in W₁ in terms of a set of linearly independent vectors. The equation a₁ - a₃ - a₄ = 0 can be rewritten as a₁ = a₃ + a₄. This means that the first component, a₁, of any vector in W₁ is determined by the third, a₃, and fourth, a₄, components. The other components, a₂ and a₅, are free variables.
Parameterizing Vectors in W₁
We can express a vector (a₁, a₂, a₃, a₄, a₅) in W₁ using the condition a₁ = a₃ + a₄. By introducing parameters for the free variables, we can rewrite any vector in W₁ in the following form:
(a₁, a₂, a₃, a₄, a₅) = (a₃ + a₄, a₂, a₃, a₄, a₅)
Now, we can express this vector as a linear combination of vectors, each corresponding to one of the free variables:
(a₃ + a₄, a₂, a₃, a₄, a₅) = a₂(0, 1, 0, 0, 0) + a₃(1, 0, 1, 0, 0) + a₄(1, 0, 0, 1, 0) + a₅(0, 0, 0, 0, 1)
Identifying the Basis for W₁
From the above expression, we can identify four vectors that span W₁:
- v₁ = (0, 1, 0, 0, 0)
- v₂ = (1, 0, 1, 0, 0)
- v₃ = (1, 0, 0, 1, 0)
- v₄ = (0, 0, 0, 0, 1)
These vectors are linearly independent, as no vector can be written as a linear combination of the others. To verify this, we can set up a linear combination equal to the zero vector and show that all the coefficients must be zero:
c₁v₁ + c₂v₂ + c₃v₃ + c₄v₄ = (0, 0, 0, 0, 0)
c₁(0, 1, 0, 0, 0) + c₂(1, 0, 1, 0, 0) + c₃(1, 0, 0, 1, 0) + c₄(0, 0, 0, 0, 1) = (0, 0, 0, 0, 0)
This gives us the following system of equations:
- c₂ + c₃ = 0
- c₁ = 0
- c₂ = 0
- c₃ = 0
- c₄ = 0
From this system, it's clear that c₁ = c₂ = c₃ = c₄ = 0, confirming the linear independence of the vectors. Therefore, the set {v₁, v₂, v₃, v₄} forms a basis for W₁.
Determining the Dimension of W₁
The dimension of a subspace is the number of vectors in its basis. Since we have found four linearly independent vectors that form a basis for W₁, the dimension of W₁ is 4. This makes sense intuitively because the original equation a₁ - a₃ - a₄ = 0 imposes one constraint on the five variables, leaving four degrees of freedom.
(b) Subspace W₂ = {(a₁, a₂, a₃, a₄, a₅) ∈ ℝ⁵ | a₂ = a₃ = a₄, a₁ + a₅ = 0}
Understanding Subspace W₂
Now, let's analyze the subspace W₂, defined by the conditions a₂ = a₃ = a₄ and a₁ + a₅ = 0. These two equations impose linear constraints on the components of the vectors in W₂, which will help us determine its basis and dimension. The conditions stipulate that the second, third, and fourth components are equal, and the sum of the first and fifth components is zero. This significantly restricts the possible vectors that can belong to W₂.
Expressing Vectors in W₂ with Parameters
To find a basis for W₂, we need to parameterize the vectors based on the given conditions. The conditions are:
- a₂ = a₃ = a₄
- a₁ + a₅ = 0, which implies a₁ = -a₅
Let's use parameters to represent these dependencies. Let a₂ = a₃ = a₄ = t and a₅ = s. Then, a₁ = -s. Therefore, any vector (a₁, a₂, a₃, a₄, a₅) in W₂ can be written as:
(-s, t, t, t, s)
Rewriting as a Linear Combination
We can rewrite this vector as a linear combination of two vectors, each corresponding to one of the parameters, s and t:
(-s, t, t, t, s) = s(-1, 0, 0, 0, 1) + t(0, 1, 1, 1, 0)
This expression reveals that any vector in W₂ can be formed by a linear combination of the vectors (-1, 0, 0, 0, 1) and (0, 1, 1, 1, 0).
Identifying the Basis for W₂
From the linear combination, we can identify two vectors that span W₂:
- u₁ = (-1, 0, 0, 0, 1)
- u₂ = (0, 1, 1, 1, 0)
To confirm that these vectors form a basis, we need to ensure they are linearly independent. We can do this by setting up a linear combination equal to the zero vector and showing that the coefficients must be zero:
c₁u₁ + c₂u₂ = (0, 0, 0, 0, 0)
c₁(-1, 0, 0, 0, 1) + c₂(0, 1, 1, 1, 0) = (0, 0, 0, 0, 0)
This gives us the following system of equations:
- -c₁ = 0
- c₂ = 0
- c₂ = 0
- c₂ = 0
- c₁ = 0
It's clear that c₁ = 0 and c₂ = 0, confirming the linear independence of the vectors. Therefore, the set {u₁, u₂} forms a basis for W₂.
Determining the Dimension of W₂
The dimension of W₂ is the number of vectors in its basis. Since we have found two linearly independent vectors that form a basis for W₂, the dimension of W₂ is 2. This dimension aligns with the number of free parameters (s and t) we introduced to describe the vectors in W₂. The two constraints, a₂ = a₃ = a₄ and a₁ + a₅ = 0, reduce the five-dimensional space to a two-dimensional subspace.
Conclusion
In summary, we have found bases and dimensions for both subspaces W₁ and W₂ of ℝ⁵. For W₁, defined by the equation a₁ - a₃ - a₄ = 0, the basis is {(0, 1, 0, 0, 0), (1, 0, 1, 0, 0), (1, 0, 0, 1, 0), (0, 0, 0, 0, 1)}, and the dimension is 4. For W₂, defined by the equations a₂ = a₃ = a₄ and a₁ + a₅ = 0, the basis is {(-1, 0, 0, 0, 1), (0, 1, 1, 1, 0)}, and the dimension is 2.
Understanding how to find bases and dimensions of subspaces is essential in linear algebra. This process involves identifying the constraints that define the subspace, expressing the vectors in terms of free parameters, and then determining the linearly independent vectors that span the space. The dimension of the subspace is simply the number of vectors in its basis. These concepts are fundamental for solving a wide range of problems in mathematics, physics, engineering, and computer science, providing a powerful toolkit for analyzing vector spaces and their properties.
