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: Set the stage.
We have $P$ of size $4\times 6$ and $Q$ of size $6\times 4$, with the products of $P$ and $Q$ being zero. This ties the range of one to the null space of the other.

Step 2: Where the ranges sit.
From $PQ=0$, every column of $Q$ is killed by $P$, so the range of $Q$ lies inside the null space of $P$. The matching condition puts the range of $P$ inside the null space of $Q$. So the containment statement holds.

Step 3: Bound the ranks.
Since the range of $P$ sits in the null space of $Q$, we get $\operatorname{rank}(P)\le \text{nullity}(Q)$. Combining the two containments with the rank nullity rule on the $4$ wide pieces forces $\operatorname{rank}(P)+\operatorname{rank}(Q)\le 4$. So that bound is true.

Step 4: The equality case.
If the range of $P$ is the whole null space of $Q$, then the inequality above becomes an equality, so $\operatorname{rank}(P)+\operatorname{rank}(Q)=4$. That statement is also true.

Step 5: The false claim.
The range of $Q$ need not fill the null space of $P$, it only sits inside it, so the equality claim there can fail.

Step 6: Conclusion.
The correct statements are the rank bound and the equality case.
\[ \boxed{B \text{ and } C} \]