Let W be the set of all vectors in V = R3 of the form (x, y, 0). Prove that W is a subspace of V = R3.

The zero vector in $\mathbb{R}^3$ is $(0, 0, 0)$. We can express this vector in the form $(x, y, 0)$ by choosing $x=0$ and $y=0$.
Since $(0, 0, 0) \in W$, $W$ is non-empty.

Let $\mathbf{u} = (x_1, y_1, 0)$ and $\mathbf{v} = (x_2, y_2, 0)$ be two arbitrary vectors in $W$.
Their sum is:

Since $x_1, x_2, y_1, y_2 \in \mathbb{R}$, it follows that $x_1 + x_2 \in \mathbb{R}$ and $y_1 + y_2 \in \mathbb{R}$.
The resulting vector $(x_1 + x_2, y_1 + y_2, 0)$ is of the form $(x, y, 0)$, so it belongs to $W$.
Thus, $W$ is closed under vector addition.

Let $\mathbf{u} = (x_1, y_1, 0)$ be a vector in $W$ and $c$ be an arbitrary scalar in $\mathbb{R}$.
Their scalar product is:

Since $c, x_1, y_1 \in \mathbb{R}$, it follows that $cx_1 \in \mathbb{R}$ and $cy_1 \in \mathbb{R}$.
The resulting vector $(cx_1, cy_1, 0)$ is of the form $(x, y, 0)$, so it belongs to $W$.
Thus, $W$ is closed under scalar multiplication.

Since all three conditions are satisfied, $W$ is a subspace of $V = \mathbb{R}^3$.
$\text{W is a subspace of } \mathbb{R}^3$