Vector Subspace Exercise 2

Let $V = \mathbb{R}^2$ be a real two-dimensional vector space, and consider the subset: $$ W = \{ (x, y) \in \mathbb{R}^2 \mid x - 2y = 3 \} $$ We want to determine whether $W$ is a vector subspace of $V$.

To do so, we must verify whether $W$ satisfies the axioms of a vector space and, more specifically, those required for a vector subspace.

Checking the Zero Vector

The defining condition for membership in $W$ is:

$$ \vec{w} = \begin{pmatrix} x \\ y \end{pmatrix} \in W \quad \Leftrightarrow \quad x - 2y = 3 $$

We begin by checking whether the zero vector is in $W$.

Substituting $x = 0$ and $y = 0$ into the equation:

$$ x - 2y = 3 \quad \Rightarrow \quad 0 - 0 = 0 \quad \neq \quad 3 $$

Since the condition is not satisfied by $(0, 0)$, the zero vector does not belong to $W$.

This violates one of the fundamental properties of vector spaces: the presence of the zero vector.

Note. Any set that fails to contain the zero vector cannot be a vector space and therefore cannot be a subspace.

As a result, there is no need to check the other subspace criteria. We can conclude immediately that $W$ is not a vector subspace of $\mathbb{R}^2$.

This completes the exercise.

And so on.