Vector Subspace Exercise 1

Let $V = \mathbb{R}^2$. We want to determine whether the following subset $W$ is a vector subspace of $V$: $$ W = \{ (x, y) \in \mathbb{R}^2 \mid 3x - 2y = 0 \} $$

To verify this, we’ll check whether $W$ satisfies the defining properties of a vector subspace.

1) The Zero Vector Belongs to W

Consider a generic vector $\vec{w} = \begin{pmatrix} x \\ y \end{pmatrix}$ in $W$. By definition,

$$ \vec{w} \in W \quad \Leftrightarrow \quad 3x - 2y = 0 $$

We check whether the zero vector is in $W$:

$$ 3(0) - 2(0) = 0 $$

Since the equation holds, the zero vector $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ belongs to $W$.

Note. It’s useful to check for the presence of the zero vector first. If it doesn’t belong to the set, then the set cannot be a subspace, and there's no need to test the other properties.

2) Closure Under Vector Addition

Let $\vec{w}_1 = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\vec{w}_2 = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$ be two arbitrary vectors in $W$. Then:

$$ 3x_1 - 2y_1 = 0 \quad \text{and} \quad 3x_2 - 2y_2 = 0 $$

We now check whether their sum also satisfies the defining condition of $W$:

$$ 3(x_1 + x_2) - 2(y_1 + y_2) = (3x_1 - 2y_1) + (3x_2 - 2y_2) = 0 + 0 = 0 $$

Hence, $\vec{w}_1 + \vec{w}_2 \in W$, so $W$ is closed under addition.

3) Closure Under Scalar Multiplication

Let $\vec{w} = \begin{pmatrix} x \\ y \end{pmatrix} \in W$, so that $3x - 2y = 0$. Let $\lambda \in \mathbb{R}$ be any scalar.

We check whether $\lambda \vec{w}$ also lies in $W$:

$$ 3(\lambda x) - 2(\lambda y) = \lambda (3x - 2y) = \lambda \cdot 0 = 0 $$

Therefore, $\lambda \vec{w} \in W$, and $W$ is closed under scalar multiplication.

Since $W$ contains the zero vector and is closed under both vector addition and scalar multiplication, $W$ is a vector subspace of $V$.

Q.E.D.