To determine whether the set $V = \{ (x, iy, 0) : x, y \in \mathbb{R} \}$ is a subspace of the vector space $\mathbb{C}^3$ over $\mathbb{C}$, we must verify if it satisfies the following three conditions that define a subspace:

1. The zero vector is in $V$:

   • The zero vector in $\mathbb{C}^3$ is $(0, 0, 0)$.
   • If we set $x = 0$ and $y = 0$ in the definition of $V$, we get $(0, i \cdot 0, 0) = (0, 0, 0)$, which is the zero vector.
   • Therefore, the zero vector is in $V$.

2. Closed under addition:

   • Let $u = (x_1, iy_1, 0)$ and $v = (x_2, iy_2, 0)$ be two vectors in $V$, where $x_1, x_2, y_1, y_2 \in \mathbb{R}$.
   • The sum of these two vectors is: $u + v = (x_1, iy_1, 0) + (x_2, iy_2, 0) = (x_1 + x_2, i(y_1 + y_2), 0)$
   • Since $x_1 + x_2 \in \mathbb{R}$ and $y_1 + y_2 \in \mathbb{R}$, the sum $u + v$ is of the form $(x, iy, 0)$, where $x \in \mathbb{R}$ and $y \in \mathbb{R}$.
   • Thus, $V$ is closed under addition.

3. Closed under scalar multiplication:

   • Let $u = (x, iy, 0) \in V$ and $\alpha \in \mathbb{C}$ be any scalar.
   • The scalar multiple of $u$ by $\alpha$ is: $\alpha u = \alpha (x, iy, 0) = (\alpha x, \alpha iy, \alpha \cdot 0) = (\alpha x, i \alpha y, 0)$
   • For $u \in V$ to remain in $V$, $\alpha x$ and $\alpha y$ must both be real numbers. However, since $\alpha \in \mathbb{C}$, this is not guaranteed, as multiplying by a complex number can produce non-real components for $\alpha x$ or $\alpha y$.
   • Therefore, $V$ is not closed under scalar multiplication by complex numbers.

Conclusion:

The set $V = \{ (x, iy, 0) : x, y \in \mathbb{R} \}$ is not a subspace of $\mathbb{C}^3$ over $\mathbb{C}$, because it is not closed under scalar multiplication by complex numbers.

Would you like more details or have any other questions? Here are 5 related questions you could explore:

1. What is the definition of a vector space and a subspace?
2. What is the difference between a real and complex vector space?
3. Can a set be a subspace of $\mathbb{R}^n$ but not $\mathbb{C}^n$?
4. How does scalar multiplication affect subspaces in different fields?
5. Can you find other examples of sets that fail to be subspaces due to scalar multiplication?

Tip: Always check closure properties (addition and scalar multiplication) when verifying if a set is a subspace of a vector space.