Linear Independence of Vectors in Higher Dimensions

Introduction to Linear Independence

When dealing with vectors in linear algebra, the concept of linear independence is fundamental. Linear independence refers to a set of vectors where none of the vectors can be expressed as a linear combination of the others. Understanding whether a set of vectors is linearly independent is crucial for various applications in mathematics and engineering, such as solving systems of linear equations and in the field of optimization. This article will focus on analyzing the linear independence of the set of vectors ((-1, 2, 5)), ((-3, 1, 11)), and ((0, -1, -2)) using various methods, particularly the determinant approach.

Method 1: Determinant Approach

To determine the linear independence of the vectors (mathbf{v_1} (-1, 2, 5)), (mathbf{v_2} (-3, 1, 11)), and (mathbf{v_3} (0, -1, -2)), we can set up a linear combination that equals the zero vector:

[c_1mathbf{v_1} c_2mathbf{v_2} c_3mathbf{v_3} mathbf{0}]

This results in the following system of equations:

[begin{align*}c_1 - 1c_2 - 3c_3 0 c_1 cdot 2 c_2 cdot 1 c_3 cdot -1 0 c_1 cdot 5 c_2 cdot 11 c_3 cdot -2 0end{align*}]

We can write this system in matrix form:

[begin{bmatrix}-1 -3 0 2 1 -1 5 11 -2end{bmatrix}begin{bmatrix}c_1 c_2 c_3end{bmatrix}begin{bmatrix}0 0 0end{bmatrix}]

To determine if there are non-trivial solutions (solutions other than (c_1 c_2 c_3 0)), we need to compute the determinant of the coefficient matrix. If the determinant is non-zero, the vectors are linearly independent.

The coefficient matrix is:

[A begin{bmatrix}-1 -3 0 2 1 -1 5 11 -2end{bmatrix}]

The determinant of the matrix (A) can be calculated as:

[ text{det}(A) -1(1 cdot -2 - (-1) cdot 11) - (-3)(2 cdot -2 - (-1) cdot 5) 0(2 cdot 11 - 1 cdot 5) ]

Calculating the determinant:

[text{det}(A) -1(-2 - (-11)) 3(-4 - (-5)) 0]

[text{det}(A) -1(9) 3(1) 0 -9 3 -6]

Since the determinant is (-6), which is non-zero, the vectors are linearly independent.

Method 2: Vector Manipulation

Another approach is to manipulate the given vectors to simplify the analysis. Let's consider the vectors ((-125, -3111, 0 - 1 - 2)). This is a more general form and can be compared with the process of finding linearly independent vectors.

We start by defining a vector (mathbf{b'} mathbf{b} - 3mathbf{a}):

[mathbf{b'} (-3, 1, 11) - 3(-1, 2, 5) (0, -5, -4)]

(mathbf{b}) and (mathbf{c}) lie in the (yz)-plane and are non-collinear since their components are not proportional. Any vector in the (yz)-plane can be expressed as a linear combination of them. Vector (mathbf{a}) has a non-zero (x)-component and thus cannot lie in the (yz)-plane. Therefore, (mathbf{a}), (mathbf{b'}), and (mathbf{c}) are linearly independent.

Since the set of vectors (mathbf{a} mathbf{b'} mathbf{c}) is derived from the original vectors (mathbf{a} mathbf{b} mathbf{c}), the original vectors must also be linearly independent.

Alternative Method: Linear Combinations and Matrix Manipulations

To further verify the linear independence, we can derive new vectors from the original set using linear combinations. For example,

[mathbf{d} -frac{1}{3}mathbf{b}mathbf{c} -frac{1}{3}(-3, 0, 9) (1, 0, -3)]

[mathbf{e} -frac{1}{5}mathbf{a}mathbf{2c}mathbf{2d} -frac{1}{5}(0, 0, -5) (0, 0, 1)]

[mathbf{f} mathbf{c}mathbf{2e} (0, -1, 0)]

[mathbf{g} mathbf{d}mathbf{3e} (1, 0, 0)]

The vectors (mathbf{e}, mathbf{f}, mathbf{g}) are linearly independent, implying that the original vectors (mathbf{a}, mathbf{b}, mathbf{c}) are also linearly independent.

Conclusion

The set of vectors ((-1, 2, 5)), ((-3, 1, 11)), and ((0, -1, -2)) are linearly independent. This conclusion is supported by multiple methods, including the determinant approach and vector manipulations. Understanding and verifying linear independence is crucial for solving complex problems in linear algebra and its applications.