(D) If $A$ is invertible, then $\det(A + B) = \det(A) + \det(B)$
Answer Key:
A, B, C
Step-by-Step Solution & Explanation
1. **Subadditivity of Rank (Option A)**: The column space of $A + B$ is contained in the sum of the column spaces of $A$ and $B$, so $\operatorname{rank}(A + B) \le \operatorname{rank}(A) + \operatorname{rank}(B)$ (True).
2. **Rank of Matrix Product (Option B)**: For any matrices $A$ and $B$, $\operatorname{rank}(AB) \le \min(\operatorname{rank}(A), \operatorname{rank}(B))$ (True).
3. **Trace Cyclicity (Option C)**: $\operatorname{trace}(AB) = \sum_i (AB)_{ii} = \sum_{i,j} A_{ij} B_{ji} = \sum_j (BA)_{jj} = \operatorname{trace}(BA)$ (True).
4. **Determinant Non-additivity (Option D)**: Determinant is not additive: $\det(A + B) \neq \det(A) + \det(B)$ in general (e.g., $A = I, B = I \implies \det(2I) = 2^n \neq 1 + 1 = 2$ for $n \ge 2$) (False).