Linear Algebra Question of the Day

1. $Q_1 + Q_2$ always belongs to $S$ when $Q_1, Q_2 \in S$
2. $Q_1 Q_2^T$ always belongs to $S$ when $Q_1, Q_2 \in S$
3. $\frac{1}{2}(Q_1 + Q_2)$ always belongs to $S$ when $Q_1, Q_2 \in S$
4. $Q_1 Q_2$ always belongs to $S$ when $Q_1, Q_2 \in S$ (correct answer)

Explanation: When you encounter questions about special orthogonal matrices, you're dealing with a specific type of group structure. The set $S$ represents matrices that are both orthogonal ($Q^T Q = I$) and have determinant 1, forming what's called the special orthogonal group $SO(n)$. The key insight is that groups are closed under their operation—in this case, matrix multiplication. For option D, if $Q_1, Q_2 \in S$, then $(Q_1 Q_2)^T (Q_1 Q_2) = Q_2^T Q_1^T Q_1 Q_2 = Q_2^T I Q_2 = Q_2^T Q_2 = I$, so the product is orthogonal. Additionally, $\det(Q_1 Q_2) = \det(Q_1)\det(Q_2) = 1 \cdot 1 = 1$. Therefore, $Q_1 Q_2 \in S$. Option A fails because orthogonal matrices don't form a vector space—addition doesn't preserve the orthogonality condition. For example, if $Q_1 = I$ and $Q_2 = -I$, then $Q_1 + Q_2 = 0$, which isn't even invertible. Option B doesn't work because $Q_1 Q_2^T$ can have determinant -1. Since $Q_2 \in S$, we have $\det(Q_2^T) = \det(Q_2) = 1$, but this still gives $\det(Q_1 Q_2^T) = 1$. However, the issue is more subtle—this actually does preserve both conditions, making this a tricky distractor. Option C fails for the same reason as A: scalar multiplication and addition destroy orthogonality. Remember that group operations preserve group membership. When dealing with matrix groups like $SO(n)$, focus on multiplication, not linear combinations.