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Suppose \(\mathbb{R}^{4}=\operatorname{Span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\\} .\) Explain why \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\\}\) is a basis for \(\mathbb{R}^{4} .\)

Short Answer

Expert verified
The set \( \{ \mathbf{v}_{1}, \ldots, \mathbf{v}_{4} \} \) is a basis because it spans \( \mathbb{R}^{4} \) and contains 4 vectors, confirming linear independence.

Step by step solution

01

Understanding the concept of a basis

A set of vectors forms a basis for a vector space if they are linearly independent and span the entire space. In this exercise, we are given that \( \mathbb{R}^{4} = \operatorname{Span}\{ \mathbf{v}_{1}, \ldots, \mathbf{v}_{4} \} \), meaning that these vectors span \( \mathbb{R}^{4} \).
02

Analyzing the dimensions

The dimension of \( \mathbb{R}^{4} \) is 4, which means we need exactly four linearly independent vectors to form a basis for it. Here, we are provided with four vectors.
03

Confirming linear independence

To confirm that \( \{ \mathbf{v}_{1}, \ldots, \mathbf{v}_{4} \} \) is a basis, we need to ensure that the vectors are linearly independent. Since they span \( \mathbb{R}^{4} \), they must also be linearly independent; otherwise, they would not span the entire space with just four vectors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Independence
In the world of vector spaces, understanding linear independence is crucial. Linear independence is a simple yet powerful property. If a set of vectors is linearly independent, it means that no vector in the set can be written as a linear combination of the others. This is a foundational requirement for a set of vectors to be considered a basis for a vector space.
To check if vectors are linearly independent, one can set up an equation where the vectors are multiplied by scalars and added together, equating the sum to zero (i.e., the zero vector). If the only solution to this equation is that all scalars are zero, then the vectors are linearly independent.
For instance, in our exercise with \( \mathbf{v}_1, \ldots, \mathbf{v}_4 \) in \( \mathbb{R}^{4} \), these vectors are indeed linearly independent. Why? Because if they weren't, we would need more than four vectors to span \( \mathbb{R}^{4} \), given that none can be a combination of the others in the span.
  • Key Point: Linear independence ensures that each vector contributes uniquely to the span of the space.
Spanning Set
A spanning set is a collection of vectors that can be combined in different ways to produce every vector in the space. Knowing whether vectors span a space helps us understand the coverage or reach of those vectors. In every spanning set of a vector space, every vector of that space can be represented as a combination of the vectors in the set.
Let's consider our exercise with \( \mathbf{v}_1, \ldots, \mathbf{v}_4 \) spanning \( \mathbb{R}^{4} \). This means that any vector in the four-dimensional real space can be created with these vectors. You practically play with combinations of these vectors to get any location in that four-dimensional space.
  • Significant Detail: \( \mathbb{R}^{4} = \text{Span}\{ \mathbf{v}_1, \ldots, \mathbf{v}_4 \} \) implies that these vectors cover the whole space.
Dimension of a Vector Space
The dimension of a vector space is essentially the number of vectors in its basis. It provides a measure of the "size" or complexity of the space. In simple terms, it's how many vectors you need to completely describe any vector in the space without redundancy.
In the case of \( \mathbb{R}^{4} \), its dimension is 4. This means you need exactly four linearly independent vectors to form a basis for it. This unique dimension dictates how many directions you can move within that space. In our exercise, having four vectors \( \mathbf{v}_1, \ldots, \mathbf{v}_4 \) matches the required dimension, confirming that they indeed form a basis.
  • Insight: The concept of dimension tells us how rich or large the space can be with minimal sets of vectors.

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Most popular questions from this chapter

Is the following difference equation of order 3\(?\) Explain. \(y_{k+3}+5 y_{k+2}+6 y_{k+1}=0\)

If a \(3 \times 8\) matrix \(A\) has rank \(3,\) find \(\operatorname{dim} \mathrm{Nul} A, \operatorname{dim} \operatorname{Row} A\) and \(\operatorname{rank} A^{T} .\)

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