Let's first summarize the problem clearly:

We are asked to check whether $V$, the set of column vectors with two real-number entries (denoted as $\left[ \begin{array}{c} x \\ y \end{array} \right]$ where $x, y \in \mathbb{R}$), is a vector space over the field of complex numbers, $\mathbb{C}$. The given operations are:

• Vector Addition: $\left[ \begin{array}{c} x \\ y \end{array} \right] + \left[ \begin{array}{c} w \\ z \end{array} \right] = \left[ \begin{array}{c} x + w \\ y + z \end{array} \right]$, where $x, y, w, z \in \mathbb{R}$.

• Scalar Multiplication: $\alpha \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} \alpha x \\ \alpha y \end{array} \right]$, where $\alpha \in \mathbb{C}$ and $x, y \in \mathbb{R}$.

Key Checks for a Vector Space:

1. Additive Closure: The sum of two vectors in $V$ must be in $V$. Since $x, y, w, z \in \mathbb{R}$, the addition $x + w$ and $y + z$ will also be real numbers, so the result will lie in $V$. Therefore, additive closure is satisfied.

2. Scalar Multiplication Closure: The scalar multiplication of a vector with a complex number $\alpha$ must result in a vector in $V$. The scalar multiplication is defined as: [ \alpha \left[ \begin{array}{c} x \ y \end{array} \right] = \left[ \begin{array}{c} \alpha x \ \alpha y \end{array} \right] ] where $x, y \in \mathbb{R}$ and $\alpha \in \mathbb{C}$. Since $x$ and $y$ are real, the multiplication by a complex number $\alpha$ can result in complex numbers. Thus, the resulting vector does not remain in $V$, as $V$ is only defined for real number entries. Therefore, scalar multiplication closure is not satisfied.

3. Other Axioms: The other axioms, such as the existence of a zero vector, the existence of additive inverses, and so on, may still hold, but scalar multiplication closure failure already disqualifies $V$ from being a vector space over $\mathbb{C}$.

Conclusion:

The problem essentially boils down to whether the closure under scalar multiplication holds, which it does not, because multiplying by a complex scalar can lead to a non-real result. Therefore, $V$ is not a vector space over the field of complex numbers.

The correct answer is: b. No, because the Additive Closure axiom is not satisfied, though the Scalar Closure axiom is satisfied.

Additional Questions:

1. Why is scalar multiplication closure an essential axiom in vector spaces?
2. How would the situation change if $V$ were defined over the field of real numbers instead of complex numbers?
3. What would need to change in the definition of scalar multiplication for $V$ to be a vector space over $\mathbb{C}$?
4. How does the closure of addition work for real-valued vector spaces?
5. Can you think of other examples where scalar multiplication fails to satisfy closure?

Tip: Always verify both closure properties (addition and scalar multiplication) for vector spaces before checking the other axioms.