Problem 7 - A Book of Abstract Algebra

Question

Prove each of the following: If \(\left\\{\mathbf{a}_{1}, \ldots, \mathbf{a}_{n}\right\\}\) is a basis of \(V\), so is \(\left\\{k_{1} \mathrm{a}_{1}, \ldots, k_{n} \mathrm{a}_{n}\right\\}\) for any nonzero scalars \(k_{1}, \ldots, k_{n}\)

Step-by-Step Solution

Answer

The set \(\{k_{1} \mathrm{a}_{1}, \ldots, k_{n} \mathrm{a}_{n}\}\) is a basis of \(V\) because it is linearly independent and spans \(V\).

Step 1: Understanding the Problem

We are given that \(\{\mathbf{a}_{1}, \ldots, \mathbf{a}_{n}\}\) is a basis of a vector space \(V\). We need to prove that if we scale each basis vector by a nonzero scalar \(k_i\), the resulting set \(\{k_{1} \mathrm{a}_{1}, \ldots, k_{n} \mathrm{a}_{n}\}\) is also a basis of \(V\).

Step 2: Recalling the Definition of a Basis

A set of vectors forms a basis of a vector space \(V\) if it is linearly independent and spans \(V\). This means any vector in \(V\) can be written as a linear combination of the basis vectors, and the only solution to the equation \(c_1\mathbf{a}_1 + \ldots + c_n\mathbf{a}_n = \mathbf{0}\) is \(c_1 = c_2 = \ldots = c_n = 0\).

Step 3: Proving Linear Independence

To show the set \(\{k_{1} \mathrm{a}_{1}, \ldots, k_{n} \mathrm{a}_{n}\}\) is linearly independent, consider the equation \(c_1(k_1 \mathbf{a}_1) + \ldots + c_n(k_n \mathbf{a}_n) = \mathbf{0}\). This simplifies to \((c_1k_1)\mathbf{a}_1 + \ldots + (c_nk_n)\mathbf{a}_n = \mathbf{0}\). Since \(\{\mathbf{a}_1, \ldots, \mathbf{a}_n\}\) is a basis, it is linearly independent, so each coefficient \(c_i k_i = 0\) implies \(c_i = 0\) for all \(i\).

Step 4: Showing the Set Spans V

Since \(\{\mathbf{a}_1, \ldots, \mathbf{a}_n\}\) spans \(V\), any vector \(\mathbf{v} \in V\) can be expressed as \(\mathbf{v} = d_1 \mathbf{a}_1 + \ldots + d_n \mathbf{a}_n\) for some scalars \(d_i\). By setting \(c_i = \frac{d_i}{k_i}\) (which is valid since \(k_ieq 0\)), \(\mathbf{v}\) can be rewritten as \(\mathbf{v} = (c_1k_1)\mathbf{a}_1 + \ldots + (c_nk_n)\mathbf{a}_n\), showing \(\{k_{1} \mathrm{a}_{1}, \ldots, k_{n} \mathrm{a}_{n}\}\) spans \(V\).

Step 5: Conclusion

The set \(\{k_{1} \mathrm{a}_{1}, \ldots, k_{n} \mathrm{a}_{n}\}\) is both linearly independent and spans \(V\), thus it is a basis of \(V\). The problem is now solved.

Key Concepts

Linear Independence

In the context of vector spaces, linear independence is a crucial concept. A set of vectors \(\{\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_n\}\) is said to be linearly independent if the sum \(c_1\mathbf{a}_1 + c_2\mathbf{a}_2 + \ldots + c_n\mathbf{a}_n = \mathbf{0}\), where \(\mathbf{0}\) is the zero vector, holds true only when \(c_1 = c_2 = \ldots = c_n = 0\), that is, all scalars are zero.
To illustrate, imagine trying to create the zero vector by combining your set of vectors. If you can only do this by multiplying each vector by zero, then your set is linearly independent.
The independence of each vector ensures no vector is redundant, meaning none of them can be written as a combination of others. This property is essential for forming a basis in a vector space.

Span

The span of a set of vectors refers to all possible vectors that can be produced by linear combinations of the set. If \(\{\mathbf{a}_1, \ldots, \mathbf{a}_n\}\) is a set of vectors, the span would be:

• All vectors \(\mathbf{v}\) that can be formed as \(d_1\mathbf{a}_1 + d_2\mathbf{a}_2 + \ldots + d_n\mathbf{a}_n\)

where \(d_1, d_2, ..., d_n\) are scalars. An entire vector space \(V\) is spanned by a set if every vector in \(V\) can be achieved as a combination of that set.
This is important when discussing a basis, because a basis not only needs to be independent, but it also needs to span the whole vector space. It ensures that every vector in the space can be represented by the basis set.

Vector Space

A vector space is a mathematical structure within which vectors can be added together and scaled by numbers, known as scalars. Vector spaces are essential in various fields of math and physics as they provide a framework for analyzing linear equations.
Properties of a vector space include:

• Closure under addition and scalar multiplication
• Associative and commutative nature of vector addition
• Existence of a zero vector and additive inverses

These properties ensure a stable base for vector operations. A basis of a vector space provides a minimal set of vectors that entirely describe the space, enabling straightforward computation and analysis.

Scalars

Scalars are the numbers used to multiply vectors in a vector space. They can be real or complex numbers, depending on the vector space considered. Scalars act as multipliers that scale the magnitude of a vector without altering its direction. For example, multiplying a vector by 2 doubles its length.
In the context of a basis, multiplying each vector by a nonzero scalar still retains the set's ability to serve as a basis because:

• The set remains linearly independent as shown when each vector's scalar can be removed from the equation independently.
• The set continues to span the vector space as rescaling does not negate any vector's reachability and coverage of the space.

Therefore, scalars play a crucial role in transforming vector sets while maintaining their structural properties as a basis.