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In the realm of linear algebra, a **vector space** is a set equipped with two operations: vector addition and scalar multiplication. These operations must satisfy certain properties to make the set a vector space. - **Vector Addition**: This operation combines any two vectors to produce another vector. - **Scalar Multiplication**: This involves multiplying a vector by a "scalar," a constant from the field over which the space is defined. For a set to qualify as a vector space, it must meet specific criteria: - Closure under addition and scalar multiplication,
meaning sums or scalar products of vectors within the set remain within the set. - Existence of an additive identity (zero vector). - Existence of additive inverses.
This ensures that for every vector in the space, there exists another vector that when added returns the zero vector. - Associativity and commutativity in addition. - Compatibility of scalar multiplication with field multiplication and identity. Such a structure is foundational in understanding a wide array of mathematical and physical systems, providing a framework for solving equations and visualizing data.

Closure under addition is one of the fundamental properties needed for a subset to be considered a subspace. This means that if you take any two vectors from the subset and add them together, the resulting vector must also be within the subset.In the step-by-step solution provided, the set of vectors in question, defined as \( U = \{(x, 0, z) \mid x, z \in \mathbb{R} \} \), is closed under addition:- Consider two vectors, \( (x_1, 0, z_1) \) and \( (x_2, 0, z_2) \) from \( U \).- Adding them yields \((x_1 + x_2, 0, z_1 + z_2)\).The sum, \((x_1 + x_2, 0, z_1 + z_2)\), is in the same form as the original vectors, exactly matching the form described in the set definition. Therefore, every sum of two vectors in \( U \) stays in \( U \). This confirms the closure under addition property, which is essential in deeming \( U \) a potential subspace of \( V \).

To determine if a subset forms a subspace within a vector space, closure under scalar multiplication must be established. This property ensures that multiplying any vector in the subset by a scalar (a real number in this context) results in another vector that still resides within the subset.In our example with the subset \( U = \{(x, 0, z) \mid x, z \in \mathbb{R} \}\), we assess scalar multiplication with any real number, \( c \):- Take a vector \( (x, 0, z) \) from \( U \).- Multiply by a scalar to get \( c(x, 0, z) = (cx, 0, cz) \).Notice that the form \((cx, 0, cz)\) follows the same pattern as the vectors in \( U \). This confirms that scalar multiplication does not move the vector outside of \( U \). Such closure under scalar multiplication is a must-have property to qualify \( U \) as a subspace of the vector space \( V \).

The **zero vector** plays a central role in the structure of a vector space. It acts as the additive identity, meaning that any vector added to the zero vector remains unchanged. What's more, the presence of the zero vector in a subset is a requirement for the subset to be recognized as a subspace.In the solution provided, we can verify the zero vector is indeed part of the subset \( U \):- If we take values \( x = 0 \) and \( z = 0 \) in the definition of \( U \),- This yields the vector \( (0, 0, 0) \), which is the zero vector.Having the zero vector in \( U \) confirms that \( U \) has the key property of an additive identity. This supports the conclusion that \( U \) is a subspace of \( V \), as it satisfies all the essential vector space axioms including closure under addition, closure under scalar multiplication, and the existence of a zero vector.