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Chapter 10: Problem 22

Let \(V\) be a finite-dimensional vector space over a field \(F\). Show that a subset \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) of \(V\) is a basis for \(V\) over \(F\) if and only if for each \(w \in V\) there exists a unique set of elements \(c_{i} \in F\) such that \(w=c_{1} v_{1}+\ldots+c_{n} v_{n}\)

Short Answer

Expert verified
A subset is a basis if and only if each vector has a unique representation as a linear combination of the basis vectors.

Step by step solution

01

Define a Basis

A subset \(\{v_1, v_2, \ldots, v_n\}\) of a vector space \(V\) is a basis if it is linearly independent and spans \(V\). This means any vector \(w \in V\) can be expressed as a linear combination of \(\{v_1, v_2, \ldots, v_n\}\).
02

Prove Sufficiency

Assume \(\{v_1, v_2, \ldots, v_n\}\) is a basis. Then each \(w \in V\) can be written as \(w = c_1v_1 + c_2v_2 + \ldots + c_nv_n\) with \(c_i \in F\). Linear independence ensures these coefficients \(c_i\) are unique. Thus, each vector in \(V\) has a unique representation in terms of the basis vectors.
03

Prove Necessity

Assume every vector \(w \in V\) can be uniquely expressed as \(w = c_1v_1 + c_2v_2 + \ldots + c_nv_n\). This means \(\{v_1, v_2, \ldots, v_n\}\) spans \(V\), as any vector \(w\) can be constructed from the elements of \(\{v_1, v_2, \ldots, v_n\}\). Uniqueness of the coefficients implies linear independence.
04

Conclude Biconditional

We have shown that \(\{v_1, v_2, \ldots, v_n\}\) is a basis if it spans \(V\) with unique coefficients, establishing the statement's biconditional. Thus, \(\{v_1, v_2, \ldots, v_n\}\) is a basis for \(V\) if and only if each \(w \in V\) has a unique representation \(w = c_1v_1 + \ldots + c_nv_n\).

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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, a linearly independent set is crucial. To understand linear independence, think of vectors as arrows pointing in space. If no vector in a set of vectors can be written as a combination of others, they are independent.
This means each vector adds a "direction" or dimension that others do not. Mathematically, if you take any set of vectors \( \{ v_1, v_2, ..., v_n \} \) and the only way to express the zero vector is by having all coefficients be zero (i.e., \( c_1v_1 + c_2v_2 + ... + c_nv_n = 0 \) implies \( c_1 = c_2 = ... = c_n = 0 \)), then the set is linearly independent.
Linear independence is essential to forming a basis, as it ensures each vector contributes uniquely to spanning the space.
Spanning Set
A spanning set of vectors essentially 'covers' the entire vector space. Imagine using a set of vectors \( \{ v_1, v_2, ..., v_n \} \) and being able to reach any point in the vector space through combinations of these vectors.
This means any vector \( w \) in the space \( V \) can be written as a linear combination of the set of vectors. Symbolically, if you can express each vector in the space as \( w = c_1v_1 + c_2v_2 + ... + c_nv_n \) where \( c_i \) are scalars from the field \( F \), then the set spans the space.
A spanning set is an integral part of defining a basis because it ensures all possible vectors in the space can be expressed using the basis vectors.
Finite-dimensional Vector Space
A finite-dimensional vector space is a set of vectors that can be described using a finite number of basis vectors. This concept is particularly important because it simplifies analysis and computation greatly.
Consider a vector space \( V \) that can be spanned by a finite set of vectors \( \{ v_1, v_2, ..., v_n \} \). The smallest possible number of vectors that can span the space is the dimension of the space.
Finite-dimensional vector spaces have a lot of structural benefits, including having well-defined bases and straightforward methods for analysis like computing dimensions, solving systems of equations, and applying transformations.
Unique Representation in Vector Spaces
Unique representation is a key feature of a basis in vector spaces. Once a vector space is defined with a basis, each vector in that space can be represented uniquely as a linear combination of the basis vectors.
This means for any vector \( w \) in the vector space \( V \), there exists only one set of scalars \( c_1, c_2, ..., c_n \) such that \( w = c_1v_1 + c_2v_2 + ... + c_nv_n \). This uniqueness is ensured by both the linear independence and spanning properties of the basis.
Consequently, unique representation is not just a mathematical nicety; it allows solutions and results to be consistent and reliable across different applications in vector space theory.

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Most popular questions from this chapter

In Exercises 13 through 17 find a basis for the indicated extension field of \(Q\) over \(Q\). $$ \mathrm{Q}(\sqrt{2}, \sqrt{3}) $$

Find the splitting field \(K\) in \(C\) of the indicated polynomial \(f(x)\) over \(Q\), and determine \([K: \mathbb{Q}]\). $$ f(x)=x^{4}+1 $$

Find a primitive element for the indicated field \(F\). $$ F=G F(9) $$

In Exercises 1 through 6 determine whether the indicated set of vectors is a basis for the indicated vector space \(V\) over the indicated field \(F\). $$ \\{(1,0,1),(0,2,2),(3,3,0)\\} \quad V=\mathbb{R}^{3} \quad F=\mathbb{R} $$

If \(E=F\left(\alpha_{1}, \ldots, \alpha_{r}\right),\) where each \(\alpha_{i}\) is algebraic over \(F\) with \(\operatorname{deg}_{F} \alpha_{i}=n_{i},\) show that \([E: F] \leq n_{1}^{*} \ldots \cdot n_{r}\)
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