Question

If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \ldots, \mathbf{x}_{k}\right\\}\) is independent, show \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{1}+\mathbf{x}_{2}, \mathbf{x}_{1}+\mathbf{x}_{2}+\mathbf{x}_{3}, \ldots, \mathbf{x}_{1}+\mathbf{x}_{2}+\cdots+\mathbf{x}_{k}\right\\}\) is also independent.

Short answer

The transformed set is independent because the only solution is zero coefficients for each vector.

Step-by-step solution

Understand Linear Independence

A set of vectors \(\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\}\) is independent if the only solution to \(c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k = \mathbf{0}\) is \(c_1 = c_2 = \cdots = c_k = 0\).

Define New Vectors

Define the new set of vectors as \(\mathbf{y}_1 = \mathbf{x}_1, \mathbf{y}_2 = \mathbf{x}_1 + \mathbf{x}_2, \ldots, \mathbf{y}_k = \mathbf{x}_1 + \mathbf{x}_2 + \cdots + \mathbf{x}_k\). We need to show that this set is independent.

Set Up the Independence Condition

Assume a linear combination \(d_1\mathbf{y}_1 + d_2\mathbf{y}_2 + \cdots + d_k\mathbf{y}_k = \mathbf{0}\). Substitute the \(\mathbf{y}_i\) with their definitions.

Substitute and Simplify

Substitute each \(\mathbf{y}_i\) with its expression: \(d_1(\mathbf{x}_1) + d_2(\mathbf{x}_1 + \mathbf{x}_2) + \cdots + d_k(\mathbf{x}_1 + \mathbf{x}_2 + \cdots + \mathbf{x}_k) = \mathbf{0}\).

Group Terms by Original Vectors

Reorganize the equation to group the same vectors: \((d_1 + d_2 + \cdots + d_k)\mathbf{x}_1 + (d_2 + \cdots + d_k)\mathbf{x}_2 + \cdots + d_k\mathbf{x}_k = \mathbf{0}\).

Conclude Independence from Zero Coefficients

Since \(\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\}\) is independent, each group coefficient must be zero: \(d_1 + d_2 + \cdots + d_k = 0\), \(d_2 + \cdots + d_k = 0\), \ldots, \(d_k = 0\). Solve these equations to show \(d_1 = 0, d_2 = 0, \ldots, d_k = 0\).

Verify Zero Coefficients Lead to Independence

The solution \(d_1 = d_2 = \cdots = d_k = 0\) confirms the set \(\{\mathbf{y}_1, \mathbf{y}_2, \ldots, \mathbf{y}_k\}\) is independent, as no non-trivial linear combination results in zero.

Key concepts

Vector Spaces

A vector space is a fundamental concept in linear algebra. It's essentially a collection of objects called vectors. These vectors can be added together and multiplied by scalars to yield another vector. The set must satisfy a few basic rules or axioms, such as closure under addition and scalar multiplication, associativity, and a zero vector.

Vector spaces provide a framework where vectors can operate within familiar arithmetic. In our exercise, the vectors \(\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\}\) reside in a vector space, adhering to the rules of vector addition and scalar multiplication.

Key characteristics include:

• **Closure**: The sum of any two vectors in the space is also within the space.
• **Associativity and Commutativity**: Vector addition is both associative and commutative.
• **Zero Vector**: There exists a zero vector that, when added to any vector, will not change it.
• **Inverse Elements**: For every vector, there exists another vector known as its additive inverse.

These properties ensure that linear combinations can indeed form other vectors within the same space.

Linear Combinations

Linear combinations are at the heart of understanding vector spaces and independence. A linear combination of a set of vectors involves multiplying each vector by a scalar and then adding the results.

In our exercise, we are tasked with proving the independence of a set derived via linear combinations. This involves checking that a particular linear combination of these results in the zero vector only when all the scalar coefficients are zero.

Here’s what you need to remember about linear combinations:

• They form the foundation for linear equations, spanning subspaces in vector space.
• They are essential for defining linear independence — if the only linear combination that results in a zero vector has all coefficients equal to zero then the vectors are independent.

By following these principles, one can understand how the new vector set maintains the structural independence.

Mathematical Proofs

Proofs are logical arguments demonstrating the truth of mathematical statements. Our task in the exercise is to prove that a specific newly defined set of vectors remains linearly independent.

Steps in a mathematical proof generally involve a clear starting assumption, application of logical reasoning, and a step-by-step demonstration leading to a conclusion.

The proofs hinge on understanding the properties and definitions—such as those of vector spaces and linear combinations. In this exercise, using known properties of the initial independent vectors, we demonstrate that no non-trivial linear combination of the new set can yield the zero vector without all coefficients being zero.

• **Assumptions**: Begin by framing hypotheses such as the given independence of original vectors.
• **Logical steps**: Utilize definitions and algebraic manipulations to achieve your goal.
• **Conclusions**: Verify that derived conditions satisfy the requirement, reinforcing the proof.

Mathematical proofs, by nature, establish certainty and clarity, ensuring that such properties are reliably understood and applied.