91Ó°ÊÓ

In a vector space, vector addition refers to the process of adding two vectors together. For a space to be closed under vector addition, the sum of any two vectors from the set must also be in the set.

Consider two vectors, say \( \begin{bmatrix} r_1 \ s_1 \ t_1 \end{bmatrix} \) and \( \begin{bmatrix} r_2 \ s_2 \ t_2 \end{bmatrix} \), both satisfying the condition \( 5r - 1 = s + 2t \). When adding these vectors, the result is \( \begin{bmatrix} r_1 + r_2 \ s_1 + s_2 \ t_1 + t_2 \end{bmatrix} \).

Plugging these into the equation:

• \( 5(r_1 + r_2) - 1 = (s_1 + s_2) + 2(t_1 + t_2) \)

simplifies using the initial conditions:

• \( 5r_1 - 1 = s_1 + 2t_1 \)
• \( 5r_2 - 1 = s_2 + 2t_2 \)

verifies that vector addition is indeed closed in this respect. However, this alone does not confirm that the set is a vector space. More properties need to be checked.

Scalar multiplication is another essential operation in vector spaces. It involves multiplying a vector by a scalar (a real number). For a set to be a vector space, it must be closed under scalar multiplication.

Take a vector \( \begin{bmatrix} r \ s \ t \end{bmatrix} \) from the set where \( 5r - 1 = s + 2t \). Now consider multiplying it by a scalar \( c \), resulting in \( \begin{bmatrix} cr \ cs \ ct \end{bmatrix} \).

The condition for closure under scalar multiplication will then be:

• \( 5(cr) - 1 = cs + 2(ct) \)

This can be rewritten using the initial condition by expanding:

• \( c(5r - 1) = c(s + 2t) \)

Hence, the multiplication by a scalar does not disrupt the equation's form. However, even if scalar closure is maintained, the presence of a zero vector is necessary for a complete vector space.

A zero vector is a vector where all components are zero, denoted as \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \). In a vector space, the zero vector plays a crucial role because it is necessary for every vector space and acts as the additive identity.

According to our given set \( W \), for the zero vector to be included, it should satisfy:

• \( 5(0) - 1 = 0 + 2(0) \), simplifying to \(-1 = 0 \), a contradiction.

This contradiction shows that the zero vector does not belong to this set, which is a clear indication that the set cannot be considered a vector space.
Therefore, missing the zero vector disqualifies the space from satisfying all vector space axioms.

Vector spaces are defined by a set of axioms that all members must follow. For any given vector space, the following conditions or axioms must be satisfied:

• Additive Closure: The sum of two vectors in the space must also be in the space.
• Scalar Multiplication Closure: Any scalar multiplied by a vector in the space should still result in a vector within the space.
• Existence of the Zero Vector: The space must contain the zero vector which acts as an identity element for addition.
• Associativity and Commutativity: Vector addition must be associative and commutative.
• Distributive Laws: Scalar distribution across vectors must satisfy certain conditions.

In the examined set, while some properties like closure under addition and scalar multiplication may hold, the absence of the zero vector violates one of these fundamental axioms. This failure means the set cannot be a vector space as not all necessary conditions are met.