91Ó°ÊÓ

To determine if a set of vectors forms a subspace, one key property to check is **closure under addition**. This ensures that if you pick any two vectors from the set and add them together, their sum is still a vector in the same set. Let's see how it applies to our given set of vectors, \( S = \left\{ \begin{pmatrix} 0 \ y \end{pmatrix} \in \mathbb{R}^2 \mid y \in \mathbb{R} \right\} \).

• Consider two vectors: \( \begin{pmatrix} 0 \ y_1 \end{pmatrix} \) and \( \begin{pmatrix} 0 \ y_2 \end{pmatrix} \).
• According to the property of addition, their sum is \( \begin{pmatrix} 0 \ y_1 + y_2 \end{pmatrix} \).
• The first component of this sum is \( 0 \), satisfying the set's condition.

Thus, the result stays within the set \( S \). No matter which vectors you choose, adding stays within the boundaries defined by the subspace. This means the set is indeed closed under addition, a necessary requirement for forming a subspace.

Another important characteristic to verify for a subspace is **closure under scalar multiplication**. This states that multiplying any vector in the set by any scalar should produce another vector still within the set.Take a vector from our set \( S \), \( \begin{pmatrix} 0 \ y \end{pmatrix} \), and a scalar \( c \). Multiplying these, we get:

• Result: \( \begin{pmatrix} 0 \ c \cdot y \end{pmatrix} \).
• The first component remains \( 0 \), adhering to the defining condition of the set.

Notice how no matter the value of \( c \) or \( y \), the product vector's first component is always \( 0 \), ensuring all results are contained within \( S \). Being closed under scalar multiplication confirms that the original definition of the set remains satisfied, establishing a crucial piece needed for a subspace.

A defining feature of any vector subspace is that it must include the **zero vector**. The zero vector serves as the identity element for both addition and scalar multiplication:

• In \( \mathbb{R}^2 \), the zero vector is \( \begin{pmatrix} 0 \ 0 \end{pmatrix} \).
• This vector naturally fits the condition \( x = 0 \) from the set \( S \).

Inclusion of the zero vector ensures that when any vector in the subspace is added to it, the vector remains unaffected, maintaining the subspace's integrity.By containing the zero vector, \( \begin{pmatrix} 0 \ 0 \end{pmatrix} \), our set \( S \) confirms this essential property of a subspace. The presence of the zero vector solidifies \( S \) as a legitimate subspace in \( \mathbb{R}^2 \), as it satisfies all necessary conditions: closure under addition, scalar multiplication, and the inclusion of the zero vector.