91Ó°ÊÓ

A vector space is a fundamental concept in Linear Algebra. It's essentially a collection of vectors that can be added together and multiplied by scalars, adhering to certain rules. In this exercise, we consider the vector space denoted as \( V = \mathbb{Q}(\sqrt{2}) \), meaning it is composed of elements like \( a + b\sqrt{2} \), where \( a \) and \( b \) are rational numbers. This indicates that our vector space is constructed by extending the rational numbers with elements that include the square root of 2. A vector space must satisfy a few conditions:

• Closure under addition and scalar multiplication.
• Contains the zero vector.
• Existence of additive inverses for each vector.

The vector space allows us to explore complex combinations of numbers even beyond our given field, in this case, \( \mathbb{Q} \). This makes the study of vector spaces a crucial part of understanding how to manipulate and comprehend more complex mathematical structures.

Linear independence is a key feature when examining whether a set of vectors can form a basis. It essentially means that no vector in the set can be written as a combination of the others. For the vectors \( \{ 1+3\sqrt{2}, 2+5\sqrt{2} \} \), we want to confirm if they are independent.
Consider a linear combination of these vectors such as \( c_1 (1 + 3\sqrt{2}) + c_2 (2 + 5\sqrt{2}) = 0 \). We equate the coefficients of independent terms to zero, forming a system of equations:

• \( c_1 + 2c_2 = 0 \)
• \( 3c_1 + 5c_2 = 0 \)

The solution \( c_1 = c_2 = 0 \) tells us there are no nontrivial solutions, establishing independence.
Linear independence ensures that our vectors are distinct and capable of spanning an entire vector space when combined. It's a critical condition for constructing a reliable basis for any vector space.

A basis is a minimal set of vectors that can be linearly combined to produce every vector in a vector space. If a set of vectors is both linearly independent and can span the vector space, then it forms a basis. For the vector space \( \mathbb{Q}(\sqrt{2}) \), which is two-dimensional over \( \mathbb{Q} \), the basis must contain exactly two linearly independent vectors.
The given set, \( \{ 1+3\sqrt{2}, 2+5\sqrt{2} \} \), meets these criteria. We already confirmed their linear independence, and since each vector uniquely contributes to the span of the space, they collectively cover every possible vector \( a + b\sqrt{2} \).

• The dimension matches the number of vectors in the set.
• They are capable of "covering" the whole space through linear combinations.

Having a basis for a vector space is crucial as it allows us to express any vector within the space uniquely as a combination of the basis vectors.

The notion of field extension is fundamental when working with vector spaces. In simple terms, a field extension allows us to expand a smaller field with additional elements, creating a larger field. Here, we have \( \mathbb{Q} \) as our original field of rational numbers. By including \( \sqrt{2} \), which doesn't exist naturally within \( \mathbb{Q} \), we establish a field extension \( \mathbb{Q}(\sqrt{2}) \).

• This extension is necessary for addressing questions that involve square roots or irrational values.
• It enables us to work within a larger framework that accommodates these elements.

This type of extension is often referred to as a degree 2 extension because the minimum polynomial for \( \sqrt{2} \) over \( \mathbb{Q} \) is of degree 2, implying that two distinct vectors/basis elements are needed to cover the extended space fully.
Field extensions are invaluable in algebra, providing a broader context to existing numerical problems and allowing solutions otherwise impossible in the original smaller field.