Question

If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \mathbf{x}_{4}, \mathbf{x}_{5}, \mathbf{x}_{6}\right\\}\) is an inde- pendent set of vectors, show that the subset \(\left\\{\mathbf{x}_{2}, \mathbf{x}_{3}, \mathbf{x}_{5}\right\\}\) is also independent.

Short answer

The subset \(\left\{\mathbf{x}_{2}, \mathbf{x}_{3}, \mathbf{x}_{5}\right\}\) is independent because they belong to a larger independent set.

Step-by-step solution

Understanding Vector Independence

A set of vectors \(\left\{ \mathbf{x_1}, \mathbf{x_2}, \ldots, \mathbf{x_6} \right\}\) is said to be independent if no vector in the set can be written as a linear combination of the others. This implies that the only solution to the equation \( c_1\mathbf{x_1} + c_2\mathbf{x_2} + \ldots + c_6\mathbf{x_6} = \mathbf{0} \) is \( c_1 = c_2 = \ldots = c_6 = 0 \).

Implying Subset Independence

Since \(\left\{ \mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \mathbf{x_4}, \mathbf{x_5}, \mathbf{x_6} \right\}\) is independent, any subset of these vectors must also be independent. This is because adding vectors to an independent set can change the dependency only if one of the added vectors can be represented as a combination of others, which does not happen here.

Confirming Independence of Subset

Specifically for the subset \(\left\{ \mathbf{x_2}, \mathbf{x_3}, \mathbf{x_5} \right\}\), the equation \( c_2\mathbf{x_2} + c_3\mathbf{x_3} + c_5\mathbf{x_5} = \mathbf{0} \) must also satisfy only the trivial solution: \( c_2 = c_3 = c_5 = 0 \). Since the vectors are part of the overall independent set, this condition holds.

Key concepts

Linear Combination

In the realm of vector spaces, the concept of a linear combination is a foundational idea. A linear combination involves creating a new vector by scaling and adding together other vectors within a set. For example, if you have vectors \( \mathbf{x_1}, \mathbf{x_2}, \ldots, \mathbf{x_6} \), a linear combination might look like this: \( c_1 \mathbf{x_1} + c_2 \mathbf{x_2} + \ldots + c_6 \mathbf{x_6} \). Here, each \( c_i \) represents a scalar that you multiply with the respective vector.The importance of linear combinations lies in their ability to express one vector in terms of others. This becomes a critical factor when determining vector independence. If any vector in a set can be expressed as a linear combination of others, then the set is not independent. However, if no such representation is possible, the set maintains independence. Thus, understanding linear combinations is pivotal for analyzing the structure and span of vector sets.

Independent Set

An independent set of vectors is one where no vector in the set can be expressed as a linear combination of the others. To verify that a set is independent, you often need to solve an equation like \( c_1\mathbf{x_1} + c_2\mathbf{x_2} + \ldots + c_n\mathbf{x_n} = \mathbf{0} \). If the only solution is \( c_1 = c_2 = \ldots = c_n = 0 \), then the set is independent. This particular solution, where all coefficients are zero, confirms that none of the vectors depend on each other in the context of linear combinations.A neat property of independent sets is consistency when reduced. If a set like \( \left\{\mathbf{x_1}, \mathbf{x_2}, \ldots, \mathbf{x_6}\right\} \) is independent, any smaller subset, such as \( \left\{\mathbf{x_2}, \mathbf{x_3}, \mathbf{x_5}\right\} \), will also be independent. This is because the removal of vectors does not introduce dependencies among the selected ones.

Trivial Solution

The concept of a trivial solution is crucial when discussing dependencies and vector independence. In the equation \( c_1\mathbf{x_1} + c_2\mathbf{x_2} + \ldots + c_n\mathbf{x_n} = \mathbf{0} \), the trivial solution is when all coefficients \( c_1, c_2, \ldots, c_n \) are zero.This solution shows that the vectors in question do not overlap or can identify as multiples of one another, meaning none of the vectors can be derived as a linear combination of others in the set.In our exercise, confirming that \( c_2 = c_3 = c_5 = 0 \) is the only way to solve \( c_2\mathbf{x_2} + c_3\mathbf{x_3} + c_5\mathbf{x_5} = \mathbf{0} \) guarantees the independence of \( \left\{\mathbf{x_2}, \mathbf{x_3}, \mathbf{x_5}\right\} \). Thus, maintaining a trivial solution is a hallmark of independent sets, separating them from dependent sets where non-zero coefficients provide alternative solutions.