Question

Consider matrices \( P \) of order \( 4 \times 6 \) and \( Q \) of order \( 6 \times 4 \) with real entries such that

• A. \( \text{rangspace}(P) \subseteq \text{nullspace}(Q) \) and \( \text{rangspace}(Q) \subseteq \text{nullspace}(P) \).
• B. \( \text{rank}(P) + \text{rank}(Q) \leq 4. \)
• C. If \( \text{rangspace}(P) = \text{nullspace}(Q) \), then \( \text{rank}(P) + \text{rank}(Q) = 4. \)
• D. \( \text{rangspace}(Q) = \text{nullspace}(P). \)

Hint:

When working with rank and nullity, always use the rank-nullity theorem to understand the relationships between the null space and range of a matrix.

Answer 1

The Correct Option is B, C

Step 1: Analyze option (A).
Since \( PQ = 0 \), the range of \( P \) lies in the null space of \( Q \). Also, since \( QP^T = 0 \), the range of \( Q \) lies in the null space of \( P \). Hence, option (A) is true.

Step 2: Analyze option (B).
The rank-nullity theorem gives that the rank of a matrix is at most the smaller of the number of rows or columns. Hence, \( \text{rank}(P) + \text{rank}(Q) \leq 4 \). Therefore, option (B) is true.

Step 3: Analyze option (C).
If \( \text{rangspace}(P) = \text{nullspace}(Q) \), then we can deduce that \( \text{rank}(P) + \text{rank}(Q) = 4 \) since the dimensions of the range and null space are equal. Hence, option (C) is true.

Step 4: Analyze option (D).
Since \( PQ = 0 \) and \( QP^T = 0 \), the rank of \( Q \) is not necessarily equal to the nullspace of \( P \). Hence, option (D) is false.

Step 5: Conclusion.
The correct answer is (B), as \( \text{rank}(P) + \text{rank}(Q) \leq 4 \).