Linear Dependence And Span Analysis Of S = {(1, 2, -1), (5, 6, -5), (3, 2, -3)} In R^3

Jul 13, 2025 by ADMIN 87 views

$Exploring Linear Dependence and Span in R^3 A Detailed Analysis of S = {(1, 2, -1), (5, 6, -5), (3, 2, -3)}$

Introduction

In the realm of linear algebra, understanding concepts like linear dependence and span is crucial for grasping the structure and properties of vector spaces. This article delves into a comprehensive analysis of a specific subset S = {(1, 2, -1), (5, 6, -5), (3, 2, -3)} within the vector space R^3. We will meticulously investigate whether this set S is linearly dependent and explore its span, ultimately demonstrating that S does not span the entirety of R^3. This exploration will not only solidify our understanding of these fundamental concepts but also provide a practical application of the theoretical framework of linear algebra.

i. Determining Linear Dependence of S

To determine if the set S is linearly dependent, we need to investigate whether there exists a non-trivial linear combination of the vectors in S that equals the zero vector. In simpler terms, we want to find scalars a, b, and c, not all zero, such that:

a(1, 2, -1) + b(5, 6, -5) + c(3, 2, -3) = (0, 0, 0)

This vector equation can be translated into a system of linear equations:

a + 5b + 3c = 0
2a + 6b + 2c = 0
-a - 5b - 3c = 0

Notice that the third equation is simply the negation of the first equation, indicating a redundancy. This suggests that the system might have infinitely many solutions, implying linear dependence. To confirm this, we can use Gaussian elimination or other methods to solve the system. Let's use Gaussian elimination by forming the augmented matrix:

[ 1  5  3 | 0 ]
[ 2  6  2 | 0 ]
[ -1 -5 -3 | 0 ]

We can perform the following row operations:

R2 -> R2 - 2R1
R3 -> R3 + R1

This yields:

[ 1  5  3 | 0 ]
[ 0 -4 -4 | 0 ]
[ 0  0  0 | 0 ]

Now, we can further simplify the second row:

R2 -> R2 / -4

[ 1  5  3 | 0 ]
[ 0  1  1 | 0 ]
[ 0  0  0 | 0 ]

Finally, we can eliminate the 5 in the first row:

R1 -> R1 - 5R2

[ 1  0 -2 | 0 ]
[ 0  1  1 | 0 ]
[ 0  0  0 | 0 ]

This reduced row echelon form corresponds to the following system of equations:

a - 2c = 0
b + c = 0

From these equations, we can express a and b in terms of c:

a = 2c
b = -c

Let c = k, where k is any scalar. Then, a = 2k and b = -k. This means we have infinitely many solutions of the form (2k, -k, k) for any scalar k. Since we can find non-trivial solutions (where not all a, b, and c are zero), the set S is linearly dependent. For example, if we let k = 1, we get the solution (2, -1, 1), which means:

2(1, 2, -1) - 1(5, 6, -5) + 1(3, 2, -3) = (0, 0, 0)

This confirms the linear dependence of S. The key to determining linear dependence lies in finding a non-trivial solution to the homogeneous system of equations derived from the linear combination of the vectors. The presence of free variables, as seen in our solution, indicates an infinite number of solutions and, therefore, linear dependence.

ii. Showing S Does Not Span R^3 and Finding Span(S)

To demonstrate that S does not span R^3, we need to show that there exists a vector in R^3 that cannot be expressed as a linear combination of the vectors in S. The definition of span(S) is the set of all possible linear combinations of the vectors in S:

span(S) = { a(1, 2, -1) + b(5, 6, -5) + c(3, 2, -3) | a, b, c ∈ R }

Let's consider an arbitrary vector (x, y, z) in R^3. We want to see if we can find scalars a, b, and c such that:

a(1, 2, -1) + b(5, 6, -5) + c(3, 2, -3) = (x, y, z)

This leads to the following system of linear equations:

a + 5b + 3c = x
2a + 6b + 2c = y
-a - 5b - 3c = z

We can form the augmented matrix:

[ 1  5  3 | x ]
[ 2  6  2 | y ]
[ -1 -5 -3 | z ]

Applying the same row operations as before (R2 -> R2 - 2R1 and R3 -> R3 + R1):

[ 1  5  3 | x ]
[ 0 -4 -4 | y - 2x ]
[ 0  0  0 | x + z ]

Further simplifying R2 (R2 -> R2 / -4):

[ 1  5  3 | x ]
[ 0  1  1 | (2x - y)/4 ]
[ 0  0  0 | x + z ]

The last row implies a crucial condition for the system to have a solution: x + z = 0. This means that not every vector (x, y, z) in R^3 can be written as a linear combination of the vectors in S. For example, the vector (1, 0, 1) does not satisfy this condition (1 + 1 ≠ 0), so it is not in the span of S. Therefore, S does not span R^3. The fact that we have a constraint on x, y, and z for a solution to exist indicates that the span of S is a subspace of R^3 with a lower dimension.

To find span(S), we recognize that the condition x + z = 0 defines a plane in R^3. We can express this plane as the set of all vectors (x, y, z) such that z = -x. Thus, the span of S consists of all vectors of the form (x, y, -x), which can be written as:

span(S) = { (x, y, -x) | x, y ∈ R } = { x(1, 0, -1) + y(0, 1, 0) | x, y ∈ R }

This shows that span(S) is a two-dimensional subspace of R^3, a plane defined by the vectors (1, 0, -1) and (0, 1, 0). These two vectors are linearly independent and form a basis for span(S). The process of determining the span involves identifying the constraints imposed by the linear combinations and expressing the resulting subspace in a convenient form. The key here was to recognize the condition x + z = 0 and use it to characterize the vectors in span(S).

Conclusion

In conclusion, our analysis has demonstrated that the set S = {(1, 2, -1), (5, 6, -5), (3, 2, -3)} is linearly dependent within R^3. This was established by finding non-trivial solutions to the homogeneous system of equations derived from the linear combination of vectors in S. Furthermore, we have rigorously shown that S does not span the entire vector space R^3. This was proven by identifying a condition (x + z = 0) that must be satisfied for a vector (x, y, z) to be within the span of S. Finally, we determined that span(S) is a two-dimensional subspace of R^3, specifically a plane defined by the equation x + z = 0, and characterized by the basis vectors (1, 0, -1) and (0, 1, 0). This comprehensive exploration has provided valuable insights into the concepts of linear dependence and span, highlighting their significance in understanding the structure of vector spaces. This understanding is vital for tackling more advanced topics in linear algebra and its applications in various fields, including engineering, physics, and computer science.

This analysis underscores the importance of understanding linear dependence and span in linear algebra. The ability to determine whether a set of vectors is linearly dependent and to characterize its span is fundamental for many applications. For instance, in computer graphics, understanding span helps in manipulating objects in 3D space. In machine learning, these concepts are crucial for dimensionality reduction and feature selection. Therefore, a strong grasp of these principles is essential for anyone working with vector spaces and linear transformations.