1. **Problem:** Prove or disprove that $M_{2\times 2}(\mathbb{R}) = \{\begin{bmatrix}a & b \\ c & d\end{bmatrix} \mid a,b,c,d \in \mathbb{R}\}$ is a vector space over $\mathbb{C}$. 2. **Recall:** A vector space over a field $\mathbb{F}$ requires scalar multiplication by elements of $\mathbb{F}$. Here, the field is $\mathbb{C}$. 3. **Check scalar multiplication:** For $M_{2\times 2}(\mathbb{R})$ to be a vector space over $\mathbb{C}$, multiplying any matrix by a complex scalar must yield a matrix still in $M_{2\times 2}(\mathbb{R})$. 4. **Test example:** Take a matrix $A = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} \in M_{2\times 2}(\mathbb{R})$ and scalar $\alpha = i \in \mathbb{C}$. Then $\alpha A = i \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}i & 0 \\ 0 & i\end{bmatrix}$. 5. **Result:** The matrix $\begin{bmatrix}i & 0 \\ 0 & i\end{bmatrix}$ has complex entries, not real. So $\alpha A \notin M_{2\times 2}(\mathbb{R})$. 6. **Conclusion:** Since scalar multiplication by complex numbers does not keep matrices in $M_{2\times 2}(\mathbb{R})$, it is **not** a vector space over $\mathbb{C}$. **Final answer:** $M_{2\times 2}(\mathbb{R})$ is not a vector space over $\mathbb{C}$.