91Ó°ÊÓ

When we talk about vectors in \( \mathbb{R}^3 \), we are referring to directed line segments that have both magnitude and direction in a three-dimensional space. Each vector can be represented as a column of three numbers, called components, usually denoted by \( x \), \( y \), and \( z \). Therefore, a typical vector in \( \mathbb{R}^3 \) looks like this: \\[\begin{bmatrix} x \ y \ z\end{bmatrix}.\] These components can be real numbers, and they correspond to the vector’s position along the x-axis, y-axis, and z-axis, respectively.
A collection of such vectors forms a vector space if it satisfies certain conditions, like having a zero vector, being closed under addition and scalar multiplication. In this exercise, we specifically examine if a particular set \( S \), with conditions \( z = 2x \) and \( y = 0 \), is a subspace of \( \mathbb{R}^3 \). Here, the vectors in \( S \) take the form \( \begin{bmatrix} x \ 0 \ 2x \end{bmatrix} \), which is a subset of vectors in \( \mathbb{R}^3 \).

Closure under addition is one of the necessary conditions for a subset to be a subspace. If you take any two vectors from the subset and add them together, their sum should still be in the subset.
For the vectors in our given subset \( S \), the general form is \( \begin{bmatrix} x \ 0 \ 2x \end{bmatrix} \). Assuming we have two vectors from this subset \( \begin{bmatrix} x_1 \ 0 \ 2x_1 \end{bmatrix} \) and \( \begin{bmatrix} x_2 \ 0 \ 2x_2 \end{bmatrix} \), adding them gives: \[\begin{bmatrix} x_1 \ 0 \ 2x_1 \end{bmatrix} + \begin{bmatrix} x_2 \ 0 \ 2x_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 \ 0 \ 2x_1 + 2x_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 \ 0 \ 2(x_1 + x_2) \end{bmatrix}.\]This result also satisfies the condition \( z = 2(x) \) and keeps \( y = 0 \). Therefore, the sum is indeed still in \( S \), confirming that \( S \) is closed under addition.

Closure under scalar multiplication is another vital property for determining if a subset is a subspace. This means that if you multiply a vector in the subset by any scalar value, the resulting vector must remain in the subset.
For our subset \( S \), the vector form is \( \begin{bmatrix} x \ 0 \ 2x \end{bmatrix} \). If you multiply this vector by a scalar \( c \), you get: \[c \cdot \begin{bmatrix} x \ 0 \ 2x \end{bmatrix} = \begin{bmatrix} cx \ 0 \ 2(cx) \end{bmatrix} = \begin{bmatrix} cx \ 0 \ 2cx \end{bmatrix}.\] This vector still satisfies both conditions specified for the subspace: \( z = 2(cx) \) and \( y = 0 \). This confirms that the set \( S \) is closed under scalar multiplication.
These two closure properties, along with the fact that the zero vector is in \( S \), demonstrate that \( S \) indeed forms a subspace of \( \mathbb{R}^3 \).