Proof: To prove that $V_C$ with the given definitions of addition and scalar multiplication is a complex vector space, we need to show that it satisfies the eight axioms of a vector space. 1. Closure under addition: Let $\mathbf{u} + i\mathbf{v}, \mathbf{u}' + i\mathbf{v}' \in V_C$. Their sum is $(\mathbf{u} + \mathbf{u}')+ i(\mathbf{v} + \mathbf{v}')$. Since $\mathbf{u}, \mathbf{v}, \mathbf{u}', \mathbf{v}' \in V$ and V is a vector space (so it's closed under addition), the sum is in $V_C$. 2. Commutativity of addition: This directly follows from the commutativity of addition in V. 3. Associativity of addition: This directly follows from the associativity of addition in V. 4. Zero vector: The zero vector of $V_C$ is $0 + i0$, which is the additive identity since for any $\mathbf{u} + i\mathbf{v} \in V_C$, we have $(\mathbf{u} + i\mathbf{v}) + (0 + i0) = \mathbf{u} + i\mathbf{v}$. 5. Additive inverses: For any $\mathbf{u} + i\mathbf{v} \in V_C$, its additive inverse is $-\mathbf{u} + i(-\mathbf{v}) = -(\mathbf{u} + i\mathbf{v})$, which is in $V_C$ since V is a vector space and has additive inverses. 6. Closure under scalar multiplication: For any complex number $c = a + bi$ and any vector $\mathbf{u} + i\mathbf{v} \in V_C$, their product is $(a\mathbf{u} - b\mathbf{v}) + i(a\mathbf{v} + b\mathbf{u})$, which is in $V_C$ since V is a vector space and is closed under scalar multiplication. 7. Distributivity of scalar multiplication with respect to vector addition and scalar addition: These properties directly follow from the distributivity of scalar multiplication in V. 8. Scalar multiplication is compatible with field multiplication: For any complex numbers $c = a + bi$, $d = e + fi$ and any vector $\mathbf{u} + i\mathbf{v} \in V_C$, we have $c(d(\mathbf{u} + i\mathbf{v})) = (cd)(\mathbf{u} + i\mathbf{v})$, which directly follows from the compatibility of scalar multiplication with field multiplication in V. 9. Identity element of scalar multiplication: For any vector $\mathbf{u} + i\mathbf{v} \in V_C$, we have $1(\mathbf{u} + i\mathbf{v}) = \mathbf{u} + i\mathbf{v}$, which directly follows from the identity element of scalar multiplication in V. Since all eight axioms hold, $V_C$ is a complex vector space. --- Expansion at 241pm Proof: The proof of the complexification of a real vector space being a complex vector space involves verifying the eight axioms of a vector space. We'll examine the proof for associativity of scalar multiplication. Associativity of Scalar Multiplication: Given (𝑎+𝑏𝑖)((𝑢+𝑖𝑣)𝑧) = ((𝑎+𝑏𝑖)(𝑢+𝑖𝑣))𝑧 for all 𝑎, 𝑏 ∈ ℝ and all 𝑢, 𝑣, 𝑧 ∈ 𝑉. We know that: (𝑎+𝑏𝑖)(𝑢+𝑖𝑣) = (𝑎𝑢−𝑏𝑣)+𝑖(𝑎𝑣+𝑏𝑢) ---(1) Now, let's multiply 𝑧 = 𝑧1 + 𝑖𝑧2 to both sides of equation (1): (𝑎+𝑏𝑖)((𝑢+𝑖𝑣)𝑧) = (𝑎𝑢−𝑏𝑣+𝑖𝑎𝑣+𝑖𝑏𝑢)(𝑧1 + 𝑖𝑧2) By distributing 𝑧1 + 𝑖𝑧2, we get: (𝑎+𝑏𝑖)((𝑢+𝑖𝑣)𝑧) = ((𝑎𝑢−𝑏𝑣)𝑧1 - 𝑎𝑣𝑧2 + 𝑏𝑢𝑧2) + 𝑖((𝑎𝑢−𝑏𝑣)𝑧2 + 𝑎𝑣𝑧1 + 𝑏𝑢𝑧1) On the other hand side, ((𝑎+𝑏𝑖)(𝑢+𝑖𝑣))𝑧 = ((𝑎𝑢−𝑏𝑣)+𝑖(𝑎𝑣+𝑏𝑢))𝑧 By distributing 𝑧1 + 𝑖𝑧2, we also get: ((𝑎+𝑏𝑖)(𝑢+𝑖𝑣))𝑧 = ((𝑎𝑢−𝑏𝑣)𝑧1 - 𝑎𝑣𝑧2 + 𝑏𝑢𝑧2) + 𝑖((𝑎𝑢−𝑏𝑣)𝑧2 + 𝑎𝑣𝑧1 + 𝑏𝑢𝑧1) Comparing both sides, we see that they are equal. Hence, (𝑎+𝑏𝑖)((𝑢+𝑖𝑣)𝑧) = ((𝑎+𝑏𝑖)(𝑢+𝑖𝑣))𝑧, proving the associativity of scalar multiplication. The other seven axioms can be proven in a similar manner. This completes the proof that $V_C$ is a complex vector space under the provided definitions of addition and scalar multiplication.