Vector Subspace Exercise 4

Let $V = \mathbb{R}^2$, and consider the subset: $$ W = \{ (x, y) \in \mathbb{R}^2 \mid x \ge 0,\ y \ge 0 \} $$ We are asked to determine whether $W$ is a vector subspace of $V$.

This set includes all points in the first quadrant of the Cartesian plane.

To qualify as a subspace, $W$ must satisfy the key properties of both a vector space and a vector subspace: it must contain the zero vector, and it must be closed under both vector addition and scalar multiplication.

1) The Zero Vector

We first check whether the zero vector belongs to $W$:

$$ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \in W $$

Since this condition is met, we cannot immediately rule out the possibility that $W$ is a subspace.

Note. Verifying the presence of the zero vector is always a good starting point. If it were not included, $W$ could not be a vector space - or a subspace - and the analysis would end here.

2) Closure Under Vector Addition

Take any two vectors in $W$:

$$ \vec{w}_1 = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}, \quad \vec{w}_2 = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}, \quad \text{with } x_1, x_2 \ge 0 \text{ and } y_1, y_2 \ge 0 $$

Their sum is:

$$ \vec{w}_1 + \vec{w}_2 = \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \end{pmatrix} $$

Since the sum of two non-negative real numbers is still non-negative, we have:

$$ x_1 + x_2 \ge 0, \quad y_1 + y_2 \ge 0 $$

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

3) Closure Under Scalar Multiplication

Now let $\vec{w} = \begin{pmatrix} x \\ y \end{pmatrix} \in W$ and $k \in \mathbb{R}$ be an arbitrary scalar.

Then:

$$ k \vec{w} = \begin{pmatrix} kx \\ ky \end{pmatrix} $$

For $k \vec{w}$ to remain in $W$, we must have:

$$ kx \ge 0, \quad ky \ge 0 $$

This condition only holds if $k \ge 0$. If $k < 0$ and $\vec{w} \ne \vec{0}$, then either $kx < 0$ or $ky < 0$, meaning $k \vec{w} \notin W$.

Therefore, $W$ is not closed under scalar multiplication, and this violates a fundamental requirement for vector subspaces.

We conclude that $W$ is not a vector subspace of $\mathbb{R}^2$.

And that completes the exercise.